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Introduction toPARTIAL DIFFERENTIAL EQUATIONS

THIRD EDITION

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New Delhi-1100012011


INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS, Third EditionK. Sankara Rao

© 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book maybe reproduced in any form, by mimeograph or any other means, without permission in writing fromthe publisher.

ISBN-978-81-203-4222-4

The export rights of this book are vested solely with the publisher.

Eleventh Printing (Third Edition) … … January, 2011

Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus,New Delhi-110001 and Printed by Syndicate Binders, A-20, Hosiery Complex, Noida, Phase-IIExtension, Noida-201305 (N.C.R. Delhi).


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Preface ix

Preface to the First and Second Edition xi

0. Partial Differential Equations of First Order 1–51

0.1 Introduction 10.2 Surfaces and Normals 20.3 Curves and Their Tangents 40.4 Formation of Partial Differential Equation 70.5 Solution of Partial Differential Equations of First Order 110.6 Integral Surfaces Passing Through a Given Curve 180.7 The Cauchy Problem for First Order Equations 210.8 Surfaces Orthogonal to a Given System of Surfaces 220.9 First Order Non-linear Equations 23

0.9.1 Cauchy Method of Characteristics 250.10 Compatible Systems of First Order Equations 330.11 Charpit’s Method 37

0.11.1 Special Types of First Order Equations 42

Exercises 49

1. Fundamental Concepts 52–105

1.1 Introduction 521.2 Classification of Second Order PDE 531.3 Canonical Forms 53

1.3.1 Canonical Form for Hyperbolic Equation 551.3.2 Canonical Form for Parabolic Equation 571.3.3 Canonical Form for Elliptic Equation 59

v

Contents


vi CONTENTS

1.4 Adjoint Operators 691.5 Riemann’s Method 711.6 Linear Partial Differential Equations with Constant Coefficiants 84

1.6.1 General Method for Finding CF of Reducible Non-homogeneousLinear PDE 86

1.6.2 General Method to Find CF of Irreducible Non-homogeneous Linear PDE 891.6.3 Methods for Finding the Particular Integral (PI) 90

1.7 Homogeneous Linear PDE with Constant Coefficients 971.7.1 Finding the Complementary Function 981.7.2 Methods for Finding the PI 99

Exercises 102

2. Elliptic Differential Equations 106–181

2.1 Occurrence of the Laplace and Poisson Equations 1062.1.1 Derivation of Laplace Equation 1062.1.2 Derivation of Poisson Equation 108

2.2 Boundary Value Problems (BVPs) 1092.3 Some Important Mathematical Tools 1102.4 Properties of Harmonic Functions 111

2.4.1 The Spherical Mean 1132.4.2 Mean Value Theorem for Harmonic Functions 1142.4.3 Maximum-Minimum Principle and Consequences 115

2.5 Separation of Variables 1222.6 Dirichlet Problem for a Rectangle 1242.7 The Neumann Problem for a Rectangle 1262.8 Interior Dirichlet Problem for a Circle 1282.9 Exterior Dirichlet Problem for a Circle 1322.10 Interior Neumann Problem for a Circle 1362.11 Solution of Laplace Equation in Cylindrical Coordinates 1382.12 Solution of Laplace Equation in Spherical Coordinates 1462.13 Miscellaneous Examples 154

Exercises 178

3. Parabolic Differential Equations 182–231

3.1 Occurrence of the Diffusion Equation 1823.2 Boundary Conditions 1843.3 Elementary Solutions of the Diffusion Equation 1853.4 Dirac Delta Function 1893.5 Separation of Variables Method 1953.6 Solution of Diffusion Equation in Cylindrical Coordinates 2083.7 Solution of Diffusion Equation in Spherical Coordinates 2113.8 Maximum-Minimum Principle and Consequences 215


CONTENTS vii

3.9 Non-linear Equations (Models) 2173.9.1 Semilinear Equations 2173.9.2 Quasi-linear Equations 2173.9.3 Burger’s Equation 2183.9.4 Initial Value Problem for Burger’s Equation 219

3.10 Miscellaneous Examples 220

Exercises 229

4. Hyperbolic Differential Equations 232–281

4.1 Occurrence of the Wave Equation 2324.2 Derivation of One-dimensional Wave Equation 2334.3 Solution of One-dimensional Wave Equation by Canonical Reduction 2364.4 The Initial Value Problem; D’Alembert’s Solution 2404.5 Vibrating String—Variables Separable Solution 2454.6 Forced Vibrations—Solution of Non-homogeneous Equation 2544.7 Boundary and Initial Value Problem for Two-dimensional Wave Equations—

Method of Eigenfunction 2574.8 Periodic Solution of One-dimensional Wave Equation in Cylindrical Coordinates 2604.9 Periodic Solution of One-dimensional Wave Equation in Spherical Polar

Coordinates 2624.10 Vibration of a Circular Membrane 2644.11 Uniqueness of the Solution for the Wave Equation 2664.12 Duhamel’s Principle 2684.13 Miscellaneous Examples 270

Exercises 279

5. Green’s Function 282–315

5.1 Introduction 2825.2 Green’s Function for Laplace Equation 2895.3 The Methods of Images 2955.4 The Eigenfunction Method 3025.5 Green’s Function for the Wave Equation—Helmholtz Theorem 3055.6 Green’s Function for the Diffusion Equation 310

Exercises 314

6. Laplace Transform Methods 316–387

6.1 Introduction 3166.2 Transform of Some Elementary Functions 3196.3 Properties of Laplace Transform 3216.4 Transform of a Periodic Function 3296.5 Transform of Error Function 3326.6 Transform of Bessel’s Function 3356.7 Transform of Dirac Delta Function 337


viii CONTENTS

6.8 Inverse Transform 3376.9 Convolution Theorem (Faltung Theorem) 3446.10 Transform of Unit Step Function 3496.11 Complex Inversion Formula (Mellin-Fourier Integral) 3526.12 Solution of Ordinary Differential Equations 3566.13 Solution of Partial Differential Equations 360

6.13.1 Solution of Diffusion Equation 3626.13.2 Solution of Wave Equation 367

6.14 Miscellaneous Examples 375

Exercises 383

7. Fourier Transform Methods 388–446

7.1 Introduction 3887.2 Fourier Integral Representations 388

7.2.1 Fourier Integral Theorem 3907.2.2 Sine and Cosine Integral Representations 394

7.3 Fourier Transform Pairs 3957.4 Transform of Elementary Functions 3967.5 Properties of Fourier Trasnform 4017.6 Convolution Theorem (Faltung Theorem) 4127.7 Parseval’s Relation 4147.8 Transform of Dirac Delta Function 4167.9 Multiple Fourier Transforms 4167.10 Finite Fourier Transforms 417

7.10.1 Finite Sine Transform 4187.10.2 Finite Cosine Transform 419

7.11 Solution of Diffusion Equation 4217.12 Solution of Wave Equation 4257.13 Solution of Laplace Equation 4287.14 Miscellaneous Examples 431

Exercises 443

Bibliography 447–448

Answers and Keys to Exercises 449–484

Index 485–488


The objective of this third edition is the same as in previous two editions: to provide a broadcoverage of various mathematical techniques that are widely used for solving and to get analyticalsolutions to Partial Differential Equations of first and second order, which occur in science andengineering. In fact, while writing this book, I have been guided by a simple teaching philosophy:An ideal textbook should teach the students to solve problems. This book contains hundreds ofcarefully chosen worked-out examples, which introduce and clarify every new concept. The corematerial presented in the second edition remains unchanged.

I have updated the previous edition by adding new material as suggested by my activecolleagues, friends and students.

Chapter 1 has been updated by adding new sections on both homogeneous and non-homogeneous linear PDEs, with constant coefficients, while Chapter 2 has been repeated as suchwith the only addition that a solution to Helmholtz equation using variables separable method isdiscussed in detail.

In Chapter 3, few models of non-linear PDEs have been introduced. In particular, the exactsolution of the IVP for non-linear Burger’s equation is obtained using Cole–Hopf function.

Chapter 4 has been updated with additional comments and explanations, for betterunderstanding of normal modes of vibrations of a stretched string.

Chapters 5–7 remain unchanged.I wish to express my gratitude to various authors, whose works are referred to while writing

this book, as listed in the Bibliography. Finally, I would like to thank all my old colleagues, friendsand students, whose feedback has helped me to improve over previous two editions.

It is also a pleasure to thank the publisher, PHI Learning, for their careful processing of themanuscript both at the editorial and production stages.

Any suggestions, remarks and constructive comments for the improvement of text are alwayswelcome.

K. SANKARA RAO

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Preface


With the remarkable advances made in various branches of science, engineering and technology,today, more than ever before, the study of partial differential equations has become essential. For,to have an in-depth understanding of subjects like fluid dynamics and heat transfer, aerodynamics,elasticity, waves, and electromagnetics, the knowledge of finding solutions to partial differentialequations is absolutely necessary.

This book on Partial Differential Equations is the outcome of a series of lectures delivered byme, over several years, to the postgraduate students of Applied Mathematics at Anna University,Chennai. It is written mainly to acquaint the reader with various well-known mathematicaltechniques, namely, the variables separable method, integral transform techniques, and Green’sfunction approach, so as to solve various boundary value problems involving parabolic, elliptic andhyperbolic partial differential equations, which arise in many physical situations. In fact, theLaplace equation, the heat conduction equation and the wave equation have been derived by takinginto account certain physical problems.

The book has been organized in a logical order and the topics are discussed in a systematicmanner. In Chapter 0, partial differential equations of first order are dealt with. In Chapter 1, theclassification of second order partial differential equations, and their canonical forms are given. Theconcept of adjoint operators is introduced and illustrated through examples, and Riemann’s methodof solving Cauchy’s problem described. Chapter 2 deals with elliptic differential equations. Also,basic mathematical tools as well as various properties of harmonic functions are discussed. Further,the Dirichlet and Neumann boundary value problems are solved using variables separable methodin cartesian, cylindrical and spherical coordinate systems. Chapter 3 is devoted to a discussion onthe solution of boundary value problems describing the parabolic or diffusion equation in variouscoordinate systems using the variables separable method. Elementary solutions are also given.Besides, the maximum-minimum principle is discussed, and the concept of Dirac delta function isintroduced along with a few properties. Chapter 4 provides a detailed study of the wave equationrepresenting the hyperbolic partial differential equation, and gives D’Alembert’s solution.

In addition, the chapter presents problems like vibrating string, vibration of a circularmembrane, and periodic solutions of wave equation, shows the uniqueness of the solutions, andillustrates Duhamel’s principle. Chapter 5 introduces the basic concepts in the construction of

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Preface to theFirst and Second Edition


Green’s function for various boundary value problems using the eigenfunction method and themethod of images. Chapter 6 on Laplace transform method is self-contained since the subjectmatter has been developed from the basic definition. Various properties of the transform andinverse transform are described and detailed proofs are given, besides presenting the convolutiontheorem and complex inversion formula. Further, the Laplace transform methods are applied tosolve several initial value, boundary value and initial boundary value problems. Finally inChapter 7, the theory of Fourier transform is discussed in detail. Finite Fourier transforms are alsointroduced, and their applications to diffusion, wave and Laplace equations have been analyzed.

The text is interspersed with solved examples; also, miscellaneous examples are given inmost of the chapters. Exercises along with hints are provided at the end of each chapter so as todrill the student in problem-solving. The preprequisites for the book include a knowledge ofadvanced calculus, Fourier series, and some understanding about ordinary differential equationsand special functions.

The book is designed as a textbook for a first course on partial differential equations for thesenior undergraduate engineering students and postgraduate students of applied mathematics,physics and engineering. The various topics covered in the book can be taught either in onesemester or in two semesters depending on the syllabi. The book would also be of interest toscientists and engineers engaged in research.

In the second edition, I have added a new chapter (Partial Differential Equations of FirstOrder). Also, some additional examples are included, which are taken from question papers forGATE in the last 10 years. This, I believe, would surely benefit students intending to appear for theGATE examination.

I am indebted to many of my colleagues in the Department of Mathematics, particularly toProf. N. Muthiyalu, Prof. Prabhamani, R. Patil, Dr. J. Pandurangan, Prof. K. Manivachakan,for their many useful comments and suggestions. I am also grateful to the authorities ofAnna University, for the encouragement and inspiration provided by them.

I wish to thank Mr. M.M. Thomas for the excellent typing of the manuscript. Besides, mygratitude and appreciation are due to the Publishers, PHI Learning, for the very careful andmeticulous processing of the manuscript, both during the editorial and production stages.

Finally, I sincerely thank my wife, Leela, daughter Aruna and son-in-law R. Parthasarathi, fortheir patience and encouragement while writing this book. I also appreciate the understandingshown by my granddaughter Sangeetha who had to forego my attention and care during the courseof my book writing.

Any constructive comments for improving the contents of this volume will be warmlyappreciated.

K. SANKARA RAO

xii PREFACE TO THE FIRST AND SECOND EDITION


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Partial differential equations of first order occur in many practical situations suchas Brownian motion, the theory of stochastic processes, radioactive disintegration, noise incommunication systems, population growth and in many problems dealing with telephonetraffic, traffic flow along a highway and gas dynamics and so on. In fact, their study isessential to understand the nature of solutions and forms a guide to find the solutions ofhigher order partial differential equations.

A first order partial differential equation (usually denoted by PDE) in two independentvariables x, y and one unknown z, also called dependent variable, is an equation of the form

, , , , 0.z z

F x y zx y

� ��

� �� (0.1)

Introducing the notation

,z

px

� ��

zq

y

��

� (0.2)

Equation (0.1) can be written in symbolic form as

( , , , , ) 0.F x y z p q � (0.3)

A solution of Eq. (0.1) in some domain Ω of 2IR is a function ( , )z f x y� defined and is

of C � in Ω should satisfy the following two conditions:

(i) For every ( , ) ,x y Ω the point ( , , , , )x y z p q is in the domain of the function F.

(ii) When ( , )z f x y� is substituted into Eq. (0.1), it should reduce to an identity in x,

y for all ( , ) .x y Ω

1

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Partial Differential Equationsof First Order


2 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

We classify the PDE of first order depending upon the form of the function F. Anequation of the form

( , , ) ( , , ) ( , , )z z

P x y z Q x y z R x y zx y

� ��

� (0.4)

is a quasi-linear PDE of first order, if the derivatives /z x� � and /z y� � that appear in the

function F are linear, while the coefficients P, Q and R depend on the independent variablesx, y and also on the dependent variable z. Similarly, an equation of the form

( , ) ( , ) ( , , )z z

P x y Q x y R x y zx y

� ��

� (0.5)

is called almost linear PDE of first order, if the coefficients P and Q are functions of the inde-pendent variables only. An equation of the form

( , ) ( , ) ( , ) ( , )z z

a x y b x y c x y z d x yx y

� ��

� (0.6)

is called a linear PDE of first order, if the function F is linear in / , /z x z y� � � � and z, while

the coefficients a, b, c and d depend only on the independent variables x and y. An equationwhich does not fit into any of the above categories is called non-linear. For example,

(i)z z

x y nzx y

� ��

�

is a linear PDE of first order.

(ii) 2z zx y z

x y

� ��

�

is an almost linear PDE of first order.

(iii) ( ) 0z z

P zx y

� ��

�

is a quasi-linear PDE of first order.

(iv)22

1z z

x y

� ��

� ��

is a non-linear PDE of first order.Before discussing various methods for finding the solutions of the first order PDEs, we

shall review some of the basic definitions and concepts needed from calculus.

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Let Ω be a domain in three-dimensional space 3IR and suppose ( , , )F x y z is a function in

the class ( ),C Ω� then the vector valued function grad F can be written as


PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER 3

grad , ,F F FFx y z

∂ ∂ ∂∂ ∂ ∂

⎛ ⎞= ⎜ ⎟⎝ ⎠

(0.7)

If we assume that the partial derivatives of F do not vanish simultaneously at any point then

the set of points (x, y, z) in ,Ω satisfying the equation

( , , )F x y z C= (0.8)

is a surface in Ω for some constant C. This surface denoted by SC is called a level surface

of F. If (x0, y0, z0) is a given point in ,Ω then by taking 0 0 0( , , ) ,F x y z C= we get an equation

of the form

0 0 0( , , ) ( , , ),F x y z F x y z= (0.9)

which represents a surface in W, passing through the point 0 0 0( , , ).x y z Here, Eq. (0.8) represents

a one-parameter family of surface in W. The value of grad F is a vector, normal to the levelsurface. Now, one may ask, if it is possible to solve Eq. (0.8) for z in terms of x and y. Toanswer this question, let us consider a set of relations of the form

1( , ),=x f u v 2 ( , ),y f u v= 3( , )z f u v= (0.10)

Here for every pair of values of u and v, we will have three numbers x, y and z, whichrepresents a point in space. However, it may be noted that, every point in space need notcorrespond to a pair u and v. But, if the Jacobian

1 2( , )0

( , )f fu v

≠∂∂ (0.11)

then, the first two equations of (0.10) can be solved and u and v, can be expressed as functionsof x and y like

( , ),u x yλ= ( , ).v x yμ=Thus, u and v are obtained once x and y are known, and the third relation of Eq. (0.10)gives the value of z in the form

3 [ ( , ), ( , )]z f x y x yλ μ= (0.12)

This is, of course, a functional relation between the coordinates x, y and z as in Eq. (0.8).Hence, any point (x, y, z) obtained from Eq. (0.10) always lie on a fixed surface. Equations(0.10) are also called parametric equations of a surface. It may be noted that the parametricequation of a surface need not be unique, which can be seen in the following example:

The parametric equationssin cos ,x r θ φ= sin sin ,y r θ φ= cosz r θ=

and2

2(1 ) cos ,(1 )

x r φ θφ

−=+

2

2(1 ) sin ,(1 )

y r −=+

φ θφ 2

21

rz φφ

=+

both represent the same surface 2 2 2 2x y z r+ + = which is a sphere, where r is a constant.



If the equation of the surface is of the form

( , )z f x y� (0.13)

Then

( , ) 0.F f x y z� � � (0.14)

Differentiating partially with respect to x and y, we obtain

0,F F z

x z x

� � ��

� � 0F F z

y z y

� � ��

�

from which we get

/

(using 0.14)/

z F x F

x F z x

� � � ��

� � �

or .F

px��

�Similarly, we obtain

Fq

y

��

� �� 1.F

z

��

� �

Hence, the direction cosines of the normal to the surface at a point (x, y, z) are given as

2 2 2 2 2 2

1, ,

1 1 1

p q

p q p q p q

� ��

(0.15)

Now, returning to the level surface given by Eq. (0.8), it is easy to write the equation of thetangent plane to the surface Sc at a point (x0, y0, z0) as

0 0 0 0 0 0 0 0 0 0 0 0( ) ( , , ) ( ) ( , , ) ( ) ( , , ) 0.F F F

x x x y z y y x y z z z x y zx y z

� � ��

� � �� (0.16)

��

A curve in three-dimensional space 3IR can be described in terms of parametric equations.

Suppose r�

denotes the position vector of a point on a curve C, then the vector equation ofC may be written as

( )r F t��

for ,t I� (0.17)

where I is some interval on the real axis. In component form, Eq. (0.17) can be written as

1( ),x f t� 2 ( ),y f t� 3( )z f t� (0.18)

where ( , , )r x y z�� and 1 2 3( ) [ ( ), ( ), ( )]F t f t f t f t��

and the functions 1 2,f f and 3f belongs to

( ).C I�



Further, we assume that

31 2 ( )( ) ( ), , (0, 0, 0)

df tdf t df t

dt dt dt� � ��

(0.19)

This non-vanishing vector is tangent to the curve C at the point (x, y, z) or at 1 2 3[ ( ), ( ), ( )]f t f t f t

of the curve C.

Another way of describing a curve in three-dimensional space 3IR is by using the factthat the intersection of two surfaces gives rise to a curve.

Let

and 1 1

2 2

( , , )

( , , )

F x y z C

F x y z C

� ��

� ��(0.20)

are two surfaces. Their intersection, if not empty, is always a curve, provided grad F1 and

grad F2 are not collinear at any point of Ω in 3IR . In other words, the intersection of surfacesgiven by Eq. (0.20) is a curve if

1 2grad ( , , ) grad ( , , ) (0, 0, 0)F x y z F x y z� � (0.21)

for every ( , , ) .x y z Ω� For various values of C1 and C2, Eq. (0.20) describes different curves.The totality of these curves is called a two parameter family of curves. Here, C1 and C2

are referred as parameters of this family. Thus, if we have two surfaces denoted by S1 and S2

whose equations are in the form

and( , , ) 0

( , , ) 0

F x y z

G x y z

� ��

� ��(0.22)

Then, the equation of the tangent plane to S1 at a point 0 0 0( , , )P x y z is

0 0 0( ) ( ) ( ) 0F F F

x x y y z zx y z

� � ��

� � � � (0.23)

Similarly, the equation of the tangent plane to S2 at the point 0 0 0( , , )P x y z is

0 0 0( ) ( ) ( ) 0.G G G

x x y y z zx y z

� � ��

� � � � (0.24)

Here, the partial derivatives / , / ,F x G x� � � � etc. are evaluated at 0 0 0( , , ).P x y z The intersection

of these two tangent planes is the tangent line L at P to the curve C, which is the intersection

of the surfaces S1 and S2. The equation of the tangent line L to the curve C at 0 0 0( , , )x y z is

obtained from Eqs. (0.23) and (0.24) as

0 0 0( ) ( ) ( )x x y y z zF G F G F G F G F G F G

y z z y z x x z x y y x

� � � � � � � � � � � ��

� � ��

� � �(0.25)



or

0 0 0( ) ( ) ( )( , ) ( , ) ( , )

( , ) ( , ) ( , )

x x y y z zF G F G F G

y z z x x y

� � ��

� � �� (0.26)

Therefore, the direction cosines of L are proportional to

( , ) ( , ) ( , ), , .

( , ) ( , ) ( , )

F G F G F G

y z z x x y

� � ��

� ��

(0.27)

For illustration, let us consider the following examples:

EXAMPLE 0.1 Find the tangent vector at (0,1, /2)π to the helix described by the equation

cos ,x t� sin ,y t� ,z t� in IR .t I �

Solution The tangent vector to the helix at (x, y, z) is

, , ( sin , cos , 1).dx dy dz

t tdt dt dt

� � � ��

We observe that the point (0,1, /2)π corresponds to /2.t π� At this point (0,1, /2),π the tangent

vector to the given helix is ( 1, 0,1).�

EXAMPLE 0.2 Find the equation of the tangent line to the space circle2 2 2 1,x y z � 0x y z �

at the point (1/ 14, 2/ 14, 3/ 14).�

Solution The space circle is described as2 2 2( , , ) 1 0

( , , ) 0

F x y z x y z

G x y z x y z

� � � � �

� � � �

Recalling Eq. (0.25), the equation of the tangent plane at (1/ 14, 2/ 14, 3/ 14)� can bewritten as

1/ 14 2/ 142 3 3 1

2 2 2 214 14 14 14

3/ 141 2

2 214 14

x y

z

� ��

��

or

1/ 14 2/ 14 3/ 14.

10/ 14 8/ 14 2/ 14

x y z� � ��



��

Suppose u and v are any two given functions of x, y and z. Let F be an arbitrary function ofu and v of the form

( , ) 0F u v � (0.28)

We can form a differential equation by eliminating the arbitrary function F. For, we differentiateEq. (0.28) partially with respect to x and y to get

0F u u F v v

p pu x z v x z � � � � � ��

� � � � � �� (0.29)

and

0F u u F v v

q qu y z v y z

� � ��

� � � �� (0.30)

Now, eliminatin /F u� � and /F v� � from Eqs. (0.29) and (0.30), we obtain

0

� ��

� �

u u v vp p

x z x z

u u v vq q

y z y z

� � � ��

which simplifies to

( , ) ( , ) ( , ).

( , ) ( , ) ( , )

u v u v u vp q

y z z x x y ��

� � �(0.31)

This is a linear PDE of the type

,Pp Qq R � (0.32)

where

( , ),

( , )

u vP

y z� ��

( , ),

( , )

u vQ

z x� ��

( , ).

( , )

u vR

x y� ��

(0.33)

Equation (0.32) is called Lagrange’s PDE of first order. The following examples illustrate theidea of formation of PDE.

EXAMPLE 0.3 Form the PDE by eliminating the arbitrary function from

(i) ( ) ( ),z f x it g x it� � where 1i � �(ii) 2 2 2( , ) 0.f x y z x y z �

Solution

(i) Given ( ) ( )z f x it g x it� � (1)



Differentiating Eq. (1) twice partially with respect to x and t, we get

2

2

( ) ( )

( ) ( ).

zf x it g x it

x

zf x it g x it

x

��

��

� ��

� ��

Here, f � indicates derivative of f with respect to ( )x it and g � indicates derivative of g

with respect to ( ).x it� Also, we have

2

2

( ) ( )

( ) ( ).

zif x it ig x it

t

zf x it g x it

t

��

��

� � ��

� � � ��

From Eqs. (2) and (3), we at once, find that

2 2

2 20,

z z

x t

� ��

� � (4)

which is the required PDE.(ii) The given relation is of the form

( , ) 0,u v� �

where 2 2 2,u x y z v x y z� � Hence, the required PDE is of the form

,Pp Qq R � (Lagrange equation) (1)

where

1 2( , )2( )

1 2( , )

1 2( , )2( )

1 2( , )

u vyu v y y

P z yzy z u v

z z

u vzu v z zQ x z

u v xz x

x x

� � � � �

� � � � �

� ��

� �

� ��

� ��

and

1 2( , )2( )

1 2( , )

u vxu v x x

R y xu v yx yy y

� � � � �

� ��

� ��



Hence, the required PDE is

2( ) 2( ) 2( )z y p x z q y x� � � �or

( ) ( ) .z y p x z q y x� � � �

EXAMPLE 0.4 Eliminate the arbitrary function from the following and hence, obtain thecorresponding partial differential equation:

(i) 2 2( )z xy f x y�

(ii) ( / ).z f xy z�

Solution

(i) Given 2 2( )z xy f x y� (1)

Differentiating Eq. (1) partially with respect to x and y, we obtain

2 2

2 2

2 ( )

2 ( )

zy xf x y p

x

zx yf x y q

y

��

��

� � � ��

� � � ��

(2)

(3)

Eliminating f � from Eqs. (2) and (3) we get

2 2 ,yp xq y x� � � (4)

which is the required PDE.

(ii) Given ( / )z f xy z� (1)

Differentiating partially Eq. (1) with respect to x and y, we get

( / )

( / )

z yf xy z p

x z

z xf xy z q

y z

��

��

� ��

� ��

(2)

(3)

Eliminating f � from Eqs. (2) and (3), we find

0xp yq� �or

px qy� (4)




EXAMPLE 0.5 Form the partial differential equation by eliminating the constants from

.z ax by ab�

Solution Given z ax by ab� (1)

Differentiating Eq. (1) partially with respect to x and y we obtain

za p

x

zb q

y

� �

� �

��

��

(2)

(3)

Substituting p and q for a and b in Eq. (1), we get the required PDE as

z px qy pq� � �

EXAMPLE 0.6 Find the partial differential equation of the family of planes, the sum ofwhose x, y, z intercepts is equal to unity.

Solution Let 1x y z

a b c� � � be the equation of the plane in intercept form, so

that 1.a b c� � � Thus, we have

11

x y z

a b a b� � �

� �(1)

Differentiating Eq. (1) with respect to x and y, we have

10

1

p

a a b� �

� �or 1

1

p

a b a� �

� �(2)

and

10

1

q

b a b� �

� �or 1

1

q

a b b� �

� �(3)

From Eqs. (2) and (3), we get

p b

q a� ��

Also, from Eqs. (2) and (4), we get

1 1p

pa a b a aq

� � � �

or

1 1.p

a pq

� � � ��



Therefore,

( )qa

p q pq=

+ −(5)

Similarly, from Eqs. (3) and (4), we find

( )

pb

p q pq=

+ -(6)

Substituting the values of a and b from Eqs. (5) and (6) respectively to Eq. (1), we have

1p q pq p q pq p q pqx y zq p pq

+ − + − + −+ + =−

or1 .x y z

q p pq p q pq+ − =

+ −

That is,

,pqpx qy zp q pq

+ − =+ −

(7)


0.5 SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

In Section 0.4, we have observed that relations of the form

( , , , , ) 0F x y z a b = (0.34)

give rise to PDE of first order of the form

( , , , , ) 0=f x y z p q (0.35)

Thus, any relation of the form (0.34) containing two arbitrary constants a and b is a solutionof the PDE of the form (0.35) and is called a complete solution or complete integral.

Consider a first order PDE of the form

( , , ) ( , , ) ( , , )z zP x y z Q x y z R x y zx y+ =∂ ∂

∂ ∂(0.36)

or simplyPp + Qq = R (0.37)

where x and y are independent variables. The solution of Eq. (0.37) is a surface S lying in

the ( , , )x y z -space, called an integral surface. If we are given that ( , )z f x y= is an integral

surface of the PDE (0.37). Then, the normal to this surface will have direction cosines



proportional to ( / , / , 1)z x z y� � � � � or ( , , 1).p q � Therefore, the direction of the normal is

given by { , , 1}.n p q� �� From the PDE (0.37), we observe that the normal n

�is perpendicular

to the direction defined by the vector { , , }t P Q R��

(see Fig. 0.1).

O

z

y

x

n� t�

( , , )x y z

Fig. 0.1 Integral surface � ( , ).z f x y

Therefore, any integral surface must be tangential to a vector with components { , , },P Q R and

hence, we will never leave the integral surface or solutions surface. Also, the total differentialdz is given by

z zdz dx dy

x y

� ��

� (0.38)

From Eqs. (0.37) and (0.38), we find

{ , , } { , , }P Q R dx dy dz� (0.39)

Now, the solution to Eq. (0.37) can be obtained using the following theorem:

Theorem 0.1 The general solution of the linear PDE

Pp Qq R �

can be written in the form ( , ) 0,F u v � where F is an arbitrary function, and 1( , , )u x y z C� and

2( , , )v x y z C� form a solution of the equation

( , , ) ( , , ) ( , , )

dx dy dz

P x y z Q x y z R x y z� � (0.40)

Proof We observe that Eq. (0.40) consists of a set of two independent ordinary differentialequations, that is, a two parameter family of curves in space, one such set can be written as



( , , )

( , , )

dy Q x y z

dx P x y z� (0.41)

which is referred to as “characteristic curve”. In quasi-linear case, Eq. (0.41) cannot be

evaluated until ( , )z x y is known. Recalling Eqs. (0.37) and (0.38), we may recast them using

matrix notation as

/

/

P Q z x R

dx dy z y dz

� ��

� � � � � ��

(0.42)

Both the equations must hold on the integral surface. For the existence of finite solutions ofEq. (0.42), we must have

0P Q P R R Q

dx dy dx dz dz dy� � � (0.43)

on expanding the determinants, we have

( , , ) ( , , ) ( , , )

dx dy dz

P x y z Q x y z R x y z� � (0.44)

which are called auxiliary equations for a given PDE.In order to complete the proof of the theorem, we have yet to show that any surface

generated by the integral curves of Eq. (0.44) has an equation of the form ( , ) 0.F u v �Let

1( , , )u x y z C� and 2( , , )v x y z C� (0.45)

be two independent integrals of the ordinary differential equations (0.44). If Eqs. (0.45)satisfy Eq. (0.44), then, we have

0u u u

dx dy dz dux y z

� � ��

� � � �

and

0.v v v

dx dy dz dvx y z

� � � ��

Solving these equations, we find

��

��

dx dyu v u v u v u v

y z z y z x x z

dzu v u v

x y y x

� � � � � � � ��

� � � ��



which can be rewritten as

( , ) ( , ) ( , )( , ) ( , ) ( , )

dx dy dzu v u v u v

y z z x x y

� ��

(0.46)

Now, we may recall from Section 0.4 that the relation ( , ) 0,F u �υ where F is an arbitrary

function, leads to the partial differential equation

( , ) ( , ) ( , )

( , ) ( , ) ( , )

u v u v u vp q

y z z x x y� ��

� � �(0.47)

By virtue of Eqs. (0.37) and (0.47), Eq. (0.46) can be written as

dx dy dz

P Q R� �

The solution of these equations are known to be 1 2( , , ) and ( , , ) .u x y z C v x y z C� � Hence,

( , ) 0F u v � is the required solution of Eq. (0.37), if u and v are given by Eq. (0.45),

We shall illustrate this method through following examples:

EXAMPLE 0.7 Find the general integral of the following linear partial differential equations:

(i) 2 ( 2 )y p xy q x z y� � �

(ii) 2 2( ) ( ) .y zx p x yz q x y� � � � �

Solution(i) The integral surface of the given PDE is generated by the integral curves of the

auxiliary equation

2 ( 2 )

dx dy dz

xy x z yy� ��

(1)

The first two members of the above equation give us

dx dy

y x��

or ,x dx y dy� �

which on integration results in

2 2

2 2

x yC� � � or 2 2

1x y C� � (2)

The last two members of Eq. (1) give

2

dy dz

y z y�

� �or 2z dy y dy y dz� � �



That is,

2 ,y dy y dz z dy� which on integration yields

22y yz C� or 2

2y yz C� � (3)

Hence, the curves given by Eqs. (2) and (3) generate the required integral surface as

2 2 2( , ) 0.F x y y yz � �

(ii) The integral surface of the given PDE is generated by the integral curves of theauxiliary equation

2 2( )

dx dy dz

y zx x yz x y� �

� � � �(1)

To get the first integral curve, let us consider the first combination as

2 2 2 2

x dx y dy dz

xy zx xy y z x y

��

� � � �

or

2 2 2 2.

( )

x dx y dy dz

z x y x y

� ��

That is,

.x dx y dy z dz �On integration, we get

2 2 2

2 2 2

x y zC� � � or 2 2 2

1x y z C � � (2)

Similarly, for getting the second integral curve, let us consider the combination such as

2 2 2 2

y dx x dy dz

y xyz x xyz x y

� ��

or

0,y dx x dy dz �which on integration results in

2xy z C� � (3)

Thus, the curves given by Eqs. (2) and (3) generate the required integral surface as

2 2 2( , ) 0.F x y z xy z � �



EXAMPLE 0.8 Use Lagrange’s method to solve the equation

0

1

x y z

z z

x y

��

α β γ

where ( , ).z z x y�

Solution The given PDE can be written as

0z z z z

x y zy x y x

� � � ��

� � �� β γ α γ α β

or

( ) ( )z z

y z z x x yx y

� ��

� � � �γ β α γ β α (1)

The corresponding auxiliary equations are

( ) ( ) ( )

dx dy dz

y z z x x yγ β α γ β α� �

� � �(2)

Using multipliers x, y, and z we find that each fraction is

.0

x dx y dy z dz� ��

Therefore,

0,x dx y dy z dz �which on integration yields

2 2 21x y z C � (3)

Similarly, using multipliers , , and ,α β γ we find from Eq. (2) that each fraction is equal to

0,dx dy dzα β γ �

which on integration gives

2x y z Cα β γ� � � (4)

Thus, the general solution of the given equation is found to be2 2 2( , ) 0F x y z x y zα β γ �



EXAMPLE 0.9 Find the general integrals of the following linear PDEs:

(i) 2 2( )pz qz z x y� �

(ii) 2 2 2( ) ( ) .x yz p y zx q z xy� � � �

Solution(i) The integral surface of the given PDE is generated by the integral curves of the

auxiliary equation

2 2( )

dx dy dz

z z z x y� ��

(1)

The first two members of Eq. (1) give

0,dx dy �which on integration yields

1x y C� � (2)

Now, considering Eq. (2) and the first and last members of Eq. (1), we obtain

221

z dzdx

z C�

�

or 221

22 ,

z dzdx

z C�

�

which on integration yields

2221ln ( ) 2z C x C� � �

or

2 22ln [ ( ) ] 2z x y x C � � (3)

Thus, the curves given by Eqs. (2) and (3) generates the integral surface for the given PDEas

2 2 2( , log { 2 } 2 ) 0F x y x y z xy x � �

(ii) The integral surface of the given PDE is given by the integral curves of the auxiliaryequation

2 2 2

dx dy dz

x yz y zx z xy� �

� � �(1)

Equation (1) can be rewritten as

( ) ( ) ( ) ( ) ( ) ( )

dx dy dy dz dz dx

x y x y z y z x y z z x x y z

� � ��

(2)



Considering the first two terms of Eq. (2) and integrating, we get

1ln ( ) ln ( ) lnx y y z C� � � �

or 1x y

Cy z

� ��

(3)

Similarly, considering the last two terms of Eq. (2) and integrating, we obtain

2y z

Cz x

� ��

(4)

Thus, the integral curves given by Eqs. (3) and (4) generate the integral surface

, 0.x y y z

Fy z z x

� ��

��

In the previous section, we have seen how a general solution for a given linear PDE can beobtained. Now, we shall make use of this general solution to find an integral surface containinga given curve as explained below:

Suppose, we have obtained two integral curves described by

1

2

( , , )

( , , )

u x y z C

v x y z C

� ��

� � (0.48)

from the auxiliary equations of a given PDE. Then, the solution of the given PDE can bewritten in the form

( , ) 0F u v � (0.49)

Suppose, we wish to determine an integral surface, containing a given curve C describedby the parametric equations of the form

( ),x x t� ( ),y y t� ( ),z z t� (0.50)

where t is a parameter. Then, the particular solution (0.48) must be like

1

2

{ ( ), ( ), ( )}

{ ( ), ( ), ( )}

�

�

u x t y t z t C

v x t y t z t C

��

��

Thus, we have two relations, from which we can eliminate the parameter t to obtain a relationof the type

1 2( , ) 0F C C � , (0.52)

which leads to the solution given by Eq. (0.49). For illustration, let us consider the followingcouple of examples.



EXAMPLE 0.10 Find the integral surface of the linear PDE2 2 2 2( ) ( ) ( )x y z p y x z q x y z � � �

containing the straight line 0, 1.x y z � �

Solution The auxiliary equations for the given PDE are

2 2 2 2( ) ( ) ( )

dx dy dz

x y z y x z x y z� �

� � � �(1)

Using the multiplier xyz, we have

0.yz dx zx dy xy dz �On integration, we get

1xyz C� (2)

Suppose, we use the multipliers x, y and z. Then find that each fraction in Eq. (1) is equalto

0,x dx y dy z dz �

which on integration yields

2 2 22x y z C � (3)

For the curve in question, we have the equations in parametric form as

,x t� ,y t� � 1z �

Substituting these values in Eqs. (2) and (3), we obtain

21

222 1

t C

t C

��

� ��(4)

Eliminating the parameter t, we find

1 21 2C C� �or

1 22 1 0C C� � �

Hence, the required integral surface is2 2 2 2 1 0.x y z xyz � �

EXAMPLE 0.11 Find the integral surface of the linear PDExp yq z� �

which contains the circle defined by2 2 2 4,x y z � 2.x y z �



Solution The integral surface of the given PDE is generated by the integral curves ofthe auxiliary equation

dx dy dz

x y z� � (1)

Integration of the first two members of Eq. (1) gives

ln ln lnx y C� or

1x

Cy� (2)

Similarly, integration of the last two members of Eq. (1) yields

2y

Cz� (3)

Hence, the integral surface of the given PDE is

, 0x y

Fy z

� ��

(4)

If this integral surface also contains the given circle, then we have to find a relation betweenx/y and y/z.

The equation of the circle is

2 2 2 4

2

x y z

x y z

� � �

� � �

(5)

(6)

From Eqs. (2) and (3), we have

1/ ,y x C� 2 1 2/ /z y C x C C� �

Substituting these values of y and z in Eqs. (5) and (6), we find

2 22

2 2 21 1 2

4,x x

xC C C

� � � or 22 2 21 1 2

1 11 4x

C C C

� ��

(7)

and

1 1 22,

x xx

C C C � or

1 1 2

1 11 2x

C C C

� ��

(8)



From Eqs. (7) and (8) we observe2

2 2 21 1 21 1 2

1 1 1 11 1 ,

C C CC C C

� ��

which on simplification gives us

21 1 2 21

2 2 20

C C C C C� � �

That is,

1 2 1 1 0.C C C� � � (9)

Now, replacing C1 by x/y and C2 by y/z, we get the required integral surface as

1 0,x y x

y z y� � �

or 1 0,x x

z y� � �

or 0.xy xz yz� � �

��

Consider an interval I on the real line. If 0 0( ), ( )x s y s and 0 ( )z s are three arbitrary functions

of a single variable s I� such that they are continuous in the interval I with their firstderivatives. Then, the Cauchy problem for a first order PDE of the form

( , , , , ) 0�F x y z p q (0.53)

is to find a region IR in ( , ),x y i.e. the space containing 0 0( ( ), ( ))x s y s for all ,�s I and a

solution ( , )φ�z x y of the PDE (0.53) such that

0 0 0[ ( ), ( )] ( )Z x s y s Z s�

and ( , )φ x y together with its partial derivatives with respect to x and y are continuous functions

of x and y in the region IR .

Geometrically, there exists a surface ( , )φ�z x y which passes through the curve ,Γ called

datum curve, whose parametric equations are

0 ( ),�x x s 0 ( ),�y y s 0 ( )�z z s

and at every point of which the direction ( , , 1)�p q of the normal is such that

( , , , , ) 0�F x y z p q

This is only one form of the problem of Cauchy.



In order to prove the existence of a solution of Eq. (0.53) containing the curve ,� wehave to make further assumptions about the form of the function F and the nature of .Γ Basedon these assumptions, we have a whole class of existence theorems which is beyond the scopeof this book. However, we shall quote one form of the existence theorem without proof,which is due to Kowalewski (see Senddon, 1986).

Theorem 0.2 If

(i) g(y) and all of its derivatives are continuous for 0| | ,δ� �y y

(ii) x0 is a given number and 0 0 0 0( ), ( )z g y q g y� � � and ( , , , )f x y z q and all of its partial

derivatives are continuous in a region S defined by

0 0 0| | , | | , | | ,δ δ δ� � � � � �x x y y q q

then, there exists a unique function ( , )φ x y such that

(a) ( , )φ x y and all of its partial derivatives are continuous in a region IR defined by

0 1| | ,δ� �x x 0 2| | ,δ� �y y

(b) For all ( , )x y in IR, ( , )φ�z x y is a solution of the equation

, , ,z z

f x y zx y

� ��

� ��

and

(c) For all values of y in the interval 0 1| | ,δ� �y y 0( , ) ( ).φ �x y g y

��

One of the useful applications of the theory of linear first order PDE is to find the systemof surfaces orthogonal to a given system of surfaces. Let a one-parameter family of surfacesis described by the equation

( , , ) �F x y z C (0.54)

Then, the task is to determine the system of surfaces which cut each of the given surfaces

orthogonally. Let ( , , )x y z be a point on the surface given by Eq. (0.54), where the normal

to the surface will have direction ratios ( / , / , / )F x F y F z� � � � � � which may be denoted by P,

Q, R.Let

( , )φ�Z x y (0.55)

be the surface which cuts each of the given system orthogonally (see Fig. 0.2).



F C= 1

F C= 2

F C= 3

Z x y = ( , )φ

Fig. 0.2 Orthogonal surface to a given system of surfaces.

Then, its normal at the point ( , , )x y z will have direction ratios ( / , / , 1)z x z y� � � � � which, of

course, will be perpendicular to the normal to the surfaces characterized by Eq. (0.54). As aconsequence we have a relation

0z z

P Q Rx y

� ��

� � (0.56)

or

�Pp Qq R (0.57)

which is a linear PDE of Lagranges type, and can be recast into

F z F z F

x x y y z

� � � � ��

� (0.58)

Thus, any solution of the linear first order PDE of the type given by either Eq. (0.57) or (0.58)is orthogonal to every surface of the system described by Eq. (0.54). In other words, thesurfaces orthogonal to the system (0.54) are the surfaces generated by the integral curves ofthe auxiliary equations

/ / /

dx dy dz

F x F y F z� � � � � �� (0.59)

��" ��#��

In this section, we will discuss the problem of finding the solution of first order non-linearpartial differential equations (PDEs) in three variables of the form

( , , , , ) 0,�F x y z p q (0.60)



where

,z

px

��

� .z

qy

��

�

We also assume that the function possesses continuous second order derivatives with respect

to its arguments over a domain Ω of ( , , , , )-space,x y z p q and either Fp or Fq is not zero at

every point such that

2 2 0.� �p qF F

The PDE (0.60) establishes the fact that at every point ( , , )x y z of the region, there exists

a relation between the numbers p and q such that ( , ) 0,φ �p q which defines the direction of

the normal { , , 1}� ��n p q to the desired integral surface ( , )�z z x y of Eq. (0.60). Thus, the

direction of the normal to the desired integral surface at certain point (x, y, z) is not defineduniquely. However, a certain cone of admissable directions of the normals exist satisfying the

relation ( , ) 0φ �p q (see Fig. 0.3).

z

y

x

O

Fig. 0.3 Cone of normals to the integral surface.

Therefore, the problem of finding the solution of Eq. (0.60) reduces to finding an integral

surface ( , ),�z z x y the normals at every point of which are directed along one of the permissible

directions of the cone of normals at that point.Thus, the integral or the solution of Eq. (0.60) essentially depends on two arbitrary

constants in the form

( , , , , ) 0,�f x y z a b (0.61)

which is called a complete integral. Hence, we get a two-parameter family of integral surfacesthrough the same point.



0.9.1 Cauchy’s Method of Characteristics

The integral surface ( , )=z z x y of Eq. (0.60) that passes through a given curve x0 = x0(s),

y0 = y0(s), z0 = z0(s) may be visualized as consisting of points lying on a certain one-parameter

family of curves ( , ), ( , ), ( , ),= = =x x t s y y t s z z t s where s is a parameter of the family called

characteristics.Here, we shall discuss the Cauchy’s method for solving Eq. (0.60), which is based

on geometrical considerations. Let ( , )=z z x y represents an integral surface S of Eq. (0.60)

in (x, y, z)-space. Then, { , , 1}−p q are the direction ratios of the normal to S. Now, the

differential equation (0.60) states that at a given point 0 0 0( , , )P x y z on S, the relationship

between p0 and q0, that is 0 0 0 0 0( , , , , ),F x y z p q need not be necessarily linear. Hence, all the

tangent planes to possible integral surfaces through P form a family of planes enveloping aconical surface called Monge Cone with P as its vertex. In other words, the problem ofsolving the PDE (0.60) is to find surfaces which touch the Monge cone at each point alonga generator. For example, let us consider the non-linear PDE

2 2 1.− =p q (0.62)

At every point of the xyz-space, the relation (0.62) can be expressed parametrically as

cosh ,μ=p sinh ,μ=q μ−∞ < < ∞ (0.63)

Then, the equation of the tangent planes at 0 0 0( , , )x y z can be written as

0 0 0( ) cosh ( )sinh ( ) 0μ μ− + − − − =x x y y z z (0.64)

Let 0 0 0( , , )P x y z be the vertex and ( , , )Q x y z be any point on the generator. Then, the direction

ratios of the generator are 0 0 0( ), ( ), ( ).− − −x x y y z z Now, the direction ratios of the axis of

the cone which is parallel to x-axis are (1, 0, 0) (see Fig. 0.4). Let the semi-verticle angle of

the cone be /4.π Then,

0 0 02 2 2

0 0 0

( )1 ( )0 ( )0 1cos4 2( ) ( ) ( )

π − + − + −= =

− + − + −

x x y y z z

x x y y z zor

2 2 2 20 0 0 0( ) ( ) ( ) 2( )− + − + − = −x x y y z z x x

or

2 2 20 0 0( ) ( ) ( ) 0− − − − − =x x y y z z (0.65)

Thus, we see that the Monge cone of the PDE (0.62) is given by Eq. (0.65). This is a right

circular cone with semi-vertical angel /4π whose axis is the straight line passing through

0 0 0( , , )x y z and parallel to z-axis.



z

y

x

O

�/4

Q x y z( , , )

P x y z( , , )0 0 0

Fig. 0.4 Monge cone.

Since an integral surface is touched by a Monge cone along its generator, we must have amethod to determine the generator of the Monge cone of the PDE (0.60) which is explainedbelow:

It may be noted that the equation of the tangent plane to the integral surface ( , )�z z x y at

the point 0 0 0( , , )x y z is given by

0 0 0( ) ( ) ( ).� � � � �p x x q y y z z (0.66)

Now, the given non-linear PDE (0.60) can be recasted into an equivalent form as

0 0 0( , , , )�q q x y z p (0.67)

indicating that p and q are not independent at 0 0 0( , , ).x y z At each point of the surface S,

there exists a Monge cone which touches the surface along the generator of the cone. Thelines of contact between the tangent planes of the integral surface and the correspondingcones, that is the generators along which the surface is touched, define a direction field onthe surface S. These directions are called the characteristic directions, also called Mongedirections on S and lie along the generators of the Monge cone. The integral curves of thisfield of directions on the integral surface S define a family of curves called characteristiccurves as shown in Fig. 0.5. The Monge cone can be obtained by eliminating p from thefollowing equations:

S

Fig. 0.5 Characteristic directions on an integral surface.



0 0 0 0 0 0( ) ( , , , ) ( ) ( )� � � � �p x x q x y z p y y z z (0.68)

and

0 0( ) ( ) 0.� � �dqx x y y

dp(0.69)

observing that q is a function of p and differentiating Eq. (0.60) with respect to p, we get

0.dF F F dq

dp p q dp

� ��

� � (0.70)

Now, eliminating ( / )dq dp from Eqs. (0.69) and (0.70), we obtain

0

0

( )0

( )

x xF F

p q y y

� ��

��

�

or

0 0� ��

p q

x x y y

F F(0.71)

Therefore, the equations describing the Monge cone are given by

0 0 0

0 0 0

0 0

( , , , ),

( ) ( ) ( )

and

.p q

q q x y z p

x x p y y q z z

x x y y

F F

��

� ��

��

The second and third of Eqs. (0.72) define the generator of the Monge cone. Solving them

for 0 0( ), ( )� �x x y y and 0( ),�z z we get

0 0 0� � ��

p q p q

x x y y z z

F F pF qF(0.73)

Finally, replacing 0 0 0( ), ( ) and ( )� � �x x y y z z by dx, dy and dz respectively, which corresponds

to infinitesimal movement from 0 0 0( , , )x y z along the generator, Eq. (0.73) becomes

.� ��p q p q

dx dy dz

F F pF qF(0.74)



Denoting the ratios in Eq. (0.74) by dt, we observe that the characteristic curves on S can beobtained by solving the ordinary differential equations

{ , , ( , ), ( , ), ( , )}pdx

F x y z x y p x y q x ydt

� (0.75)

and

{ , , ( , ), ( , ), ( , )}.qdy

F x y z x y p x y q x ydt

� (0.76)

Also, we note that

dz z dx z dy dx dyp q

dt x dt y dt dt dt

� ��

� �

Therefore,

� �p qdz

pF qFdt

(0.77)

Along the characteristic curve, p is a function of t, so that

dp p dx p dy

dt x dt y dt

� ��

�

Now, using Eqs. (0.75) and (0.76), the above equation becomes

.dp p F p F

dt x p y q� � � � ��

Since �xy yxz z or ,�y xp q we have

dp p F q F

dt x p x q

� � � ��

� (0.78)

Also, differentiating Eq. (0.60) with respect to x, we find

0F F F p F q

px z p x q x

� � � � � ��

� (0.79)

Using Eq. (0.79), Eq. (0.78) becomes

( )� � �x zdp

F pFdt

(0.80)

Similarly, we can show that

( )� � �y zdq

F qFdt

(0.81)



Thus, given an integral surface, we have shown that there exists a family of characteristiccurves along which x, y, z, p and q vary according to Eqs. (0.75), (0.76), (0.77), (0.80) and(0.81). Collecting these results together, we may write

,

,

,

( ) and

( ).

��

� ��

p

q

p q

x z

y z

dxF

dt

dyF

dt

dzpF qF

dt

dpF pF

dt

dqF qF

dt

��

These equations are known as characteristic equations of the given PDE (0.60). The last threeequations of (0.82) are also called compatibility conditions. Without knowing the solution

( , )�z z x y of the PDE (0.60), it is possible to find the functions ( ), ( ), ( ), ( ), ( )x t y t z t p t q t from

Eqs. (0.82). That is, we can find the curves ( ), ( ), ( )� � �x x t y y t z z t called characteristics

and at each point of a characteristic, we can find the numbers ( )�p p t and ( )�q q t that

determine the direction of the plane

( ) ( ) ( ).p X x q Y y Z z+ =� � � (0.83)

The characteristics, together with the plane (0.83) referred to each of its points is called a

characteristic strip. The solution ( ), ( ), ( ), ( ), ( )� � � � �x x t y y t z z t p p t q q t of the

characteristic equations (0.82) satisfy the strip condition

( ) ( )� �dz dx dyp t q t

dt dt dt(0.84)

It may be noted that not every set of five functions can be interpreted as a strip. A strip should

satisfy that the planes with normals ( , , 1)p q � be tangential to the characteristic curve. That is, they

must satisfy the strip condition (0.84) and the normals should vary continuously along thecurve.

An important consequence of the Cauchy’s method of characteristic is stated in thefollowing theorem.

Theorem 0.3 Along every strip (characteristic strip) of the PDE: ( , , , , ) 0,�F x y z p q the

function ( , , , , )F x y z p q is constant.



Proof Along the characteristic strip, we have

{ ( ), ( ), ( ), ( ), ( )} � � � � �d F dx F dy F dz F dp F dqF x t y t z t p t q t

dt x dt y dt z dt p dt q dt

� � � � ��

Now, using the results listed in Eq. (0.82), the right-hand side of the above equation becomes

( ) ( ) ( ) 0. � � �x p y q z p q p x z q y zF F F F F pF qF F F pF F F qF

Hence, the function ( , , , , )F x y z p q is constant along the strip of the characteristic equations

of the PDE defined by Eq. (0.60).For illustration, we consider the following examples:

EXAMPLE 0.12 Find the characteristics of the equation �pq z and determine the integral

surface which passes through the straight line 1, .� �x z y

Solution If the initial data curve is given in parametric form as

0 ( ) 1,�x s 0 ( ) ,�y s s 0 ( ) ,�z s s

then ordinarily the solution is sought in parametric form as

( , ),�x x t s ( , ),�y y t s ( , ).�z z t s

Thus, using the given data, the differential equation becomes

0 0( ) ( ) 0� � �p s q s s F (1)

and the strip condition gives

0 01 (0) (1)� �p q or 0 1.�q (2)

Therefore,

0 1,�q 0 �p s (unique initial strip) (3)

Now, the characteristic equations for the given PDE are

,�dxq

dt,�dy

pdt

2 ,�dzpq

dt,�dp

pdt

�dqq

dt(4)

On integration, we get

1 2 2 3

1 4 1 2 5

exp ( ), exp ( ), exp ( )

exp ( ) , 2 exp (2 )

� � � ��

� � ��

p c t q c t x c t c

y c t c z c c t c(5)

Now, taking into account the initial conditions

0 0 0 0 01, , , , 1� � � � �x y s z s p s q (6)



we can determine the constants of integration and obtain (since 2 31, 0)� �c c

exp ( ), exp ( ), exp ( )

exp ( ), exp (2 )

� � � ��

� � ��

p s t q t x t

y s t z s t(7)

Consequently, the required integral surface is obtained from Eq. (7) as

.�z xy

EXAMPLE 0.13 Find the characteristics of the equation �pq z and hence, determine the

integral surface which passes through the parabola 20, .� �x y z

Solution The initial data curve is

20 0 0( ) 0, ( ) , ( ) .� � �x s y s s z s s

Using this data, the given PDE becomes

20 0( ) ( ) 0� � �p s q s s F (1)

The strip condition gives

0 02 (0) (1)� �s p q or 0 2 0� �q s (2)

Therefore,

0 2�q s and 20 0 0/ /2

2� � � s

p z q s s (3)

Now, the characteristic equations of the given PDE are given by

, , 2 , ,� � � � �dx dy dz dp dqq p pq p q

dt dt dt dt dt(4)

On integration, we obtain

1 2 2 3

1 4 1 2 5

exp ( ), exp ( ), exp ( )

exp ( ) , exp (2 )

� � � ��

� � ��

p c t q c t x c t c

y c t c z c c t c(5)

Taking into account the initial conditions2

0 0 0 0 00, , , /2, 2 ,� � � � �x y s z s p s q s

we find

1 2 3 4 5/2, 2 , 2 , /2, 0� � � � � �c s c s c s c s c



Therefore, we have

2

exp ( ), 2 exp ( ),2

2 [exp ( ) 1], [exp ( ) 1]2

exp (2 )

��

� ��

sp t q s t

sx s t y t

z s t

��

Eliminating s and t from the last three equations of (6), we get216 (4 ) .� z y x

This is the required integral surface.

EXAMPLE 0.14 Find the characteristics of the PDE

2 2 2 �p q

and determine the integral surface which passes through 0, .� �x z y

Solution The initial data curve is

0 0 0( ) 0, ( ) , ( ) .� � �x s y s s z s s

Using this data, the given PDE becomes

2 20 0 2 0� � � �p q F (1)

and the strip condition gives

0 01 (0) (1)� �p q or 0 1 0� �q (2)

Hence,

0 01, 1� � �q p (3)

Now, the characteristic equations for the given PDE are given by

2 22 , 2 , 2 2 4

0, 0

��

dx dy dzp q p q

dt dt dt

dp dq

dt dt

(4)


1 2 1 3

2 4 5

, , 2

2 , 4

� � � ��

� � ��

p c q c x c t c

y c t c z t c��



Taking into account the initial conditions

0 0 0 0 00, , , 1, 1,� � � � �x y s z s p q

we find

1, 1, 2 ,

2 , 4 .

p q x t

y t s z t s

� � � � � ��

� � � � ��(6)

The last-three equations of (6) are parametric equations of the desired integral surface. Eliminatingthe parameters s and t, we get

.� �z y x

��

Two first order PDEs are said to be compatible, if they have a common solution. We shallnow derive the necessary and sufficient conditions for the two partial differential equations

( , , , , ) 0�f x y z p q (0.85)

and

( , , , , ) 0�g x y z p q (0.86)

to be compatible.

Let ( , )0

( , )

f gJ

p q

��

� � (0.87)

Since Eqs. (0.85) and (0.86) have common solution, we can solve them and obtainexplicit expressions for p and q in the form

( , , ), ( , , )φ ψ� �p x y z q x y z (0.88)

and then, the differential relation

�p dx q dy dz

or

( , , ) ( , , )φ ψ �x y z dx x y z dy dz (0.89)

should be integrable, for which the necessary condition is

curl 0X X� ��

where { , , 1}.X � �φ ψ That is,

ˆ ˆˆˆˆ ˆ( ) / / / 0

1

ji ki j k x y z� � � � � ��

�

φ ψφ ψ



or

( ) ( )φ ψ ψ φ ψ φ� � �z z x y


ψ φψ φ ψφ � x z y z (0.90)

Now, differentiating Eq. (0.85) with respect to x and z, we get

0x p qp q

f f fx x

� ��

� � �

and

0z p qp q

f f fz z

� ��

� � �

But, from Eq. (0.89), we have

,p

x x

� ��

� φ q

x x

� ��

� ψ and so on.

Using these results, the above equations can be recast into

0φ ψ �x p x q xf f f

and

0.φ ψ �z p z q zf f f

Multiplying the second one of the above pair by φ and adding to the first one, we readily

obtain

( ) ( ) ( ) 0φ φ φφ ψ φψ �x z p x z q x zf f f f

Similarly, from Eq. (0.86) we can deduce that

( ) ( ) ( ) 0x z p x z q x zg g g gφ φ φφ ψ φψ �

Solving the above pair of equations for ( ),ψ φψ�x z we have

( ) 1 1

( ) ( )

ψ φψφ φ

� �

� �x z

p x z p x z q p q pf g g g f f f g g f J

or

1[( ) ( )]

1 ( , ) ( , )

( , ) ( , )

x z p x p x p z p zf g g f f g g fJ

f g f g

J x p z p

� ��

� � � � �

� ��

ψ φψ φ

φ (0.91)



where J is defined in Eq. (0.87). Similarly, differentiating Eq. (0.85) with respect to y and zand using Eq. (0.88), we can show that

1 ( , ) ( , )

( , ) ( , )y zf g f g

J y q z qφ ψφ ψ

� � � � �

� �

� ��

(0.92)

Finally, substituting the values of ψ φψ�x z and φ ψφy z from Eqs. (0.91) and (0.92) into

Eq. (0.90), we obtain

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

f g f g f g f g

x p z p y q z q

� � � ��

� � � � �

� �φ ψ

In view of Eqs. (0.88), we can replace φ and ψ by p and q, respectively to get

( , ) ( , ) ( , ) ( , )0

( , ) ( , ) ( , ) ( , )

f g f g f g f gp q

x p z p y q z q

� � � ��

� (0.93)

This is the desired compatibility condition. For illustration, let us consider the followingexamples:

EXAMPLE 0.15 Show that the following PDEs

� �xp yq x �� 2x p q xz �are compatible and hence, find their solution.

Solution Suppose, we have

0,� � � �f xp yq x (1)

and

2 0.g x p q xz� � � (2)

Then,

2 2 2 2 22

22

( 1)( , )2 ,

(2 )( , )

( , ) 0 ,( , )

pf g x px x x p xz xz x p xxp zx p x

f g x xz p x x

��

��

��

�

� ��

( , ),

( , ) 0 1

( , ) 0 .( , ) 1

q yf gq

y q

yf gxy

z q x

� ��

��

�

��

��



and we find

2 2 2

2

2

( , ) ( , ) ( , ) ( , )

( , ) ( , ) ( , ) ( , )

( )

0

f g f g f g f gp q xz x p x px q qxy

x p z p y q z q

xz q qxy x

xz q x qy x

xz q x p

� � � ��

� � � � � � � � �

� � � �

� � � �

� � �

�

Hence, the given PDEs are compatible.Now, solving Eqs. (1) and (2) for p and q, we obtain

3 2 2

1� �

� � �p q

xyz x x x z x x y

from which we get

(1 ) 1

(1 ) 1

x yz yzp

x xy xy

� �

and2 ( ) ( )

(1 ) 1

x z x x z xq

x xy xy

� ��

In order to get the solution of the given system, we have to integrate Eq. (0.89), that is

(1 ) ( )

1 1

��

yz x z xdz dx dy

xy xy(3)

or

( ) ( )

1 1

� ��

y z x x z xdz dx dx dy

xy xy

or

1

� ��

dz dx y dx x dy

z x xy


ln ( ) ln (1 ) ln .z x xy c� � That is,

(1 )z x c xy� � Hence, the solution of the given system is found to be

(1 ),� z x c xy (4)

which is of one-parameter family.



�� )��

In this section, we will discuss a general method for finding the complete integral or completesolution of a nonlinear PDE of first order of the form

( , , , , ) 0.�f x y z p q (0.94)

This method is known as Charpit’s method. The basic idea in Charpit’s method is the introductionof another PDE of first order of the form

( , , , , ) 0�g x y z p q (0.95)

and then, solve Eqs. (0.94) and (0.95) for p and q and substitute in

( , , , ) ( , , , ) .� dz p x y z a dx q x y z a dy (0.96)

Now, the solution of Eq. (0.96) if it exists is the complete integral of Eq. (0.94).The main task is the determination of the second equation (0.95) which is already discussed

in the previous section. Now, what is required, is to seek an equation of the form

( , , , , ) 0�g x y z p q

compatible with the given equation

( , , , , ) 0�f x y z p q

for which the necessary and sufficient condition is

( , ) ( , ) ( , ) ( , )0.

( , ) ( , ) ( , ) ( , )

f g f g f g f gp q

x p z p y q z q

� � � ��

� (0.97)

On expansion, we have

0

f g f g f g f gp

x p p x z p p z

f g f g f g f gq

y q q y z q q z

� � � � � � � ��

� � � � � � � ��

� ��

� � � ��

which can be recast into

( ) ( ) ( ) 0.p q p q x z y zg g g g g

f f pf qf f pf f qfx y z p q � � ��

� � � � ��

This is a linear PDE, from which we can determine g. The auxiliary equations of (0.98) are

( )

( )

p q p q x z

y z

dx dy dz dp

f f pf qf f pf

dq

f qf

� � ��

��

(0.99)



These equations are called Charpit’s equations. Any integral of Eq. (0.99) involving p or qor both can be taken as the second relation (0.95). Then, the integration of Eq. (0.96) givesthe complete integral as desired. It may be noted that all charpits equations need not be used,but it is enough to choose the simplest of them. This method is illustrated through thefollowing examples:

EXAMPLE 0.16 Find the complete integral of2 2( ) �p q y qz

Solution Suppose

2 2( ) 0� � �f p q y qz (1)

then, we have

2 20, ,x y zf f p q f q� � � � �

2 , 2 ,p qf py f qy z� � �

Now, the charpits auxiliary equations are given by

( ) ( )p q p q x z y z

dx dy dz dp dq

f f pf qf f pf f qf� � � �

� � � � �

That is,

2 2

2 2 2

2 2 2 2

[( ) ]

� ��

� ��

dx dy dz

py qy z p y q y qz

dp dq

pq p q q(2)

From the last two members of Eq. (2), we have

2��

dp dq

pq p

or

0 �p dp q dq


2 2 (constant) �p q a (3)



From Eqs. (1) and (3), we obtain

22 2 2 2

0 or /

and

( )/

� � � ��

� � ��

ay qz q ay z

ayp a az a y z

z

(4)

Substituting these values of p and q in

,dz p dx q dy� we get

2 2 2az a y aydz dx dy

z z

��

or

2 2 2� � �z dz ay dy az a y dx

which can be rewritten as2 2 2 1/2( )d az a y

dxa

� �

On integration, we find

2 2 2az a yx b

a

��

or2 2 2( ) ( / )x b z a y � �

Hence, the complete integral is2 2 2( ) / . �x b y z a

EXAMPLE 0.17 Find the complete integral of the PDE:2 .�z pq xy

Solution In this example, given

2 .� �f z pq xy (1)

Then, we have

, , 2

, .

� � � � �

� � � �

x y z

p q

f pqy f pqx f z

f qxy f pxy



Now, the Charpit’s auxiliary equations are given by

.( ) ( )

� � � �� p q p q x z y z

dx dy dz dp dq

f f pf qf f pf f qf

That is,

2 2 2� � � �

� � � � �dx dy dz dp dq

qxy pxy pq xy pqy pz pqx qz(2)

From Eq. (2), it follows that

/ / / /

2 2� � �

� � � �dp p dq q dx x dy y

qy z px z qy px


/ / / /� � ��

dp p dq q dx x dy y

qy px qy px

or

� � �dp dq dy dx

p q y x


(constant)p x

cq y

�

or

/�p cqy x

From the given PDE, we have2 2 2� �z pq xy cq y

which gives2 2 2/�q z cy or / / ,q z cy az y� �

where 1/ .�a c

Hence,

/ .�p z ax


,� dz p dx q dy

we get

� z azdz dx dy

ax y



or

1� dz dx dya

z a x y


1ln ln ln ln� � �z x a y b

a

or1/a az bx y�

which is the complete integral of the given PDE.

EXAMPLE 0.18 Find the complete integral of2 2 2 2 4 0 � �x p y q

using Charpit’s method.

Solution The Charpit’s equations for the given PDE can be written as

2 2 2 2 2 2 2

2

2 2 2( ) 2

2

� � ��

��

dx dy dz dp

x p y q x p y q xp

dq

yq(1)

Considering the first and last but one of Eq. (1), we have

2 22 2��

dx dp

x p xpor 0 �dx dp

x p


ln ( ) ln�xp a or �xp a (2)

From the given PDE and using the result (2), we get

2 2 24� �y q a (3)

Substituting one set of p and q values from Eqs. (2) and (3) in

,� dz p dx q dy

we find that

24 .� �dx dydz a a

x y

On integration, the complete integral of the given PDE is found to be

2ln 4 ln .� � � �z a x a y b



�� 2+&1$3� �(2+*� -/� �10*,�0.+0� �4%$,1-5*

Type I Equations Involving p and q only.

That is, equations of the type

( , ) 0.�f p q (0.100)

Let 0� �z ax by c is a solution of the given PDE, described by ( , ) 0,�f p q then

,z

p ax

��

� � .z

q by

��

� �

Substituting these values of p and q in the given PDE, we get

( , ) 0�f a b (0.101)

Solving for b, we get, ( ),φ�b a say. Then,

( )φ� z ax a y c (0.102)

is the complete integral of the given PDE.

EXAMPLE 0.19 Find a complete integral of the equation

1.p q �

Solution The given PDE is of the form ( , ) 0.�f p q Therefore, let us assume the solution

in the form

� z ax by c

where

1 �a b or 2(1 )� �b a

Hence, the complete integral is found to be2(1 ) .� � z ax a y c

EXAMPLE 0.20 Find the complete integral of the PDE

1.�pq

Solution Since the given PDE is of the form ( , ) 0,�f p q we assume the solution in

the form ,� z ax by c where 1�ab or 1/ .�b a Hence, the complete integral is

1.� � �z ax y c

a

Type II Equations Not Involving the Independent Variables.


( , , ) 0�f z p q (0.103)



As a trial solution, let us assume that z is a function of ,� �u x ay where a is an arbitraryconstant. Then,

( ) ( )� � z f u f x ay (0.104)

� � � �

� � � �

z dz u dzp

x du x du

z dz u dzq a

y du y du

� ��

Substituting these values of p and q in the given PDE, we get

, , 0� � ��

dz dzf z a

du du(0.105)

which is an ordinary differential equation of first order.

Solving Eq. (0.105) for / ,dz du we obtain

( , ) (say)φ�dzz a

duor

.( , )φ

�dzdu

z a


( , )φ� � dz

u cz a

That is,

( , ) � � F z a u c x ay c (0.106)


EXAMPLE 0.21 Find the complete integral of

(1 ) �p q qz

Solution Let us assume the solution in the form

( )� � z f u x ay

Then,

,� dzp

du� dz

q adu

Substituting these values in the given PDE, we get

1.

� �� dz dz dz

a azdu du du



That is,

1� �dza az

duor

1�

�dz

a duaz


ln ( 1)� � � az u c x ay c

which is the required complete integral.

EXAMPLE 0.22 Find the complete integral of the PDE:2 2 2 1. �p z q

Solution Let us assume that ( )� � z f u x ay is a solution of the given PDE. Then,

,� dzp

du� dz

q adu

Substituting these values of p and q in the given PDE, we obtain

2 22 2 1

� � � �� dz dz

z adu du

That is,

22 2( ) 1

� � � �� dz

z adu

or2 2

1�

�

dz

du z a

or

2 2� �z a dz du


2 2 2 2 2

ln2 2

� � � ��

z z a a z z ax ay b

a

which is the required complete integral of the given PDE.

Type III Separable Equations.

An equation in which z is absent and the terms containing x and p can be separated from thosecontaining y and q is called a separable equation.


( , ) ( , )�f x p F y q (0.107)

As a trial solution, let us assume that

( , ) ( , ) (say)� �f x p F y q a (0.108)



Now, solving them for p and q, we obtain

( ),φ�p x ( )ψ�q y

Since

z zdz dx dy p dx q dy

x y

� ��

� �

or

( ) ( )φ ψ� dz x dx y dy

On integration, we get the complete integral in the form

( ) ( )φ ψ� � �� z x dx y dy b (0.109)

EXAMPLE 0.23 Find the complete integral of the PDE:2 2 2(1 ) �p y x qx

Solution The given PDE is of separable type and can be rewritten as

2 2

2

(1 ) � �p x qa

yx(say), an arbitrary constant.

Then,

2,

1�

axp

x�q ay


,� dz p dx q dy

we get

21�

axdz dx ay dy

xOn integration, we obtain

2 212

� � � �az a x y b


EXAMPLE 0.24 Find the complete integral of2 2 � p q x y

Solution The given PDE is of separable type and can be rewritten as2 2 (say)� � � �p x y q a



Then,

,� �p x a .� q y a

Now, substituting these values of p and q in

dz p dx q dy� we find

dz x a dx y a dy�

On integration, the complete integral is found to be

3/2 3/22 2( ) ( ) .

3 3� � � � �z x a y a b

Type IV Clairaut’s Form

A first order PDE is said to be of Clairaut’s form if it can be written as

( , )� z px qy f p q (0.110)

The corresponding Charpit’s equations are

� �

� ��

p q p q

dx dy dz

x f y f px qy pf qf

dp dq

p p q q(0.111)

The integration of the last two equations of (0.111) gives us

,�p a �q b

Substituting these values of p and q in the given PDE, we get the required complete integralin the form

( , )� z ax by f a b (0.112)

EXAMPLE 0.25 Find the complete integral of the equation

2 21� z px qy p q

Solution The given PDE is in the Clairaut’s form. Hence, its complete integral is

2 21 .� z ax by a b

EXAMPLE 0.26 Find the complete integral of

( ) ( ) 1 � � �p q z xp yq



Solution The given PDE can be rewritten as

1�

z xp yqp q

which is in the Clairaut’s form,

( , )� z px qy f p q

Hence, the complete integral of the given PDE is

1.� � �

�z ax by

a b

EXAMPLE 0.27 Solve the following Cauchy IVP:

PDE: ut + cux = 0, x � R, t > 0

IC: u(x, 0) = f(x), x � R.

Solution The characteristic equations for the given problem is given by 1 0

dt dx du

c� � .

On integration, we get: u = constant and x – ct = � (constant). This linear problemhas a unique solution, given by u(x, t) = f(x – ct).

This is a right travelling wave with speed c, of course with no change in shape. Thecharacteristic line x = � + ct, gives rise to a system of parallel, straight lines in the (x, t)-planeas shown in Fig. 0.6.

Fig. 0.6 The family of characteristic lines: x – ct = �.

We observe that one of these lines that passes through the point (x, t) intersects thex-axis at (�, 0) moving with speed c, having a slope –1/c.

EXAMPLE 0.28 In classical mechanics, the Hamilton–Jacobi equation for the problem ofone-dimensional, Harmonic oscillator is given by the differential equation as (see–SankaraRao, 2005).



221 1

02 2

S SKq

m q t

� � � � � �� ,

where S = S(p, q, t), p = S

q

and K is a constant. Using Charpits method, find S.

Solution Following the notation of Eq. (0.94) we rewrite

221 1

( , , , , )2 2t q

S Sf t q S S S Kq

m q t

� � � � � �� (1)

which gives us

0, , 0, 1,t q

qt q S S S

Sf f Kq f f f

m� � � � � .

Then, the Charpits auxiliary equations (0.99) assumes the following form:

21 / 0/

qt

q t q

dSdSdt dq dS

S m KqS S m� � � �

��(2)

Considering the second and last members, we have /

q

q

dSdq

S m Kq��

.


221

2 2qS

Kq am� � (constant of integration).

Equation (1) then becomes St = –a,

and 2 22q

aS Km q

K� ��

.

Substituting St and Sq into

dS = St dt + Sq dqand integrating, we arrive at

1/222a

S at Km q dq CK

� �� or 2 2 1/2( )S at Km q dq C�� where �2 =

2a

K and C is another constant of integration.



�6��

1. Eliminate the arbitrary function in the following and hence obtain the correspondingPDE

( ).z x y f xy�

2. Form the PDE from the following by eliminating the constants

2 2( ) ( ).z x a y b�

3. Find the integral surface (general solution) of the differential equation

2 2 ( ) .z z

x y x y zx y

� ��

�

4. Find the general integrals of the following linear PDEs:

(i)2

2y zp xzq y

x� �

(ii) ( 1) ( 1) .y p x q z �

5. Find the integral surface of the linear PDE

� �xp yq z

which contains the circle 2 2 1, 1.x y z � �6. Find the equation of the integral surface of the PDE

2 ( 3) (2 ) (2 3)� � � �y z p x z q y x

which contains the circle 2 2 2 , 0. � �x y x z

7. Find the general integral of the PDE

( ) ( )� � � �x y p y x z q z

which contains the circle 2 2 1, 1. � �x y z

8. Find the solution of the equation

2 21( ) ( ) ( )

2� � � � �z p q p x q y

which passes through the x-axis.9. Find the characteristics of the equation

�pq xy

and determine the integral surface which passes through the curve , 0.� �z x y

10. Determine the characteristics of the equation2 2� �z p q

and find the integral surface which passes through the parabola 24 0, 0. � �z x y



11. Show that the PDEs

�xp yq and ( ) 2 �z xp yq xy

are compatible and hence find its solution.12. Show that the equations

2 2 1 �p q and 2 2( )� �p q x pz

are compatible and hence find its solution.13. Find the complete integral of the equation

2 2( ) �p q x pz

where / , / .p z x q z y� � � �� 14. Find the complete integrals of the equations

(i) 5 3 2 24 6 2 0� � �px q x x z

(ii) 22( ) . �z xp yq yp

15. Find the complete integral of the equation

.� �p q pq

16. Find the complete integrals of the following equations:(i) � �z pq p q

(ii) 2 2 2 2 2 2 2 2( ). � p q x y x q x y

17. Find the complete integral of the PDE

sin ( )� �z px qy pq

18. Find the complete integrals of the following PDEs:

(i) 3 2 2 3 3 3 2 2( ) 0xp q yp q p q zp q � �

(ii) 2 2 2 2( ) ( ).p qz p xq p q yp q�

19. Find the surface which intersects the surfaces of the system

( ) (3 1) � z x y c z

orthogonally and passes through the circle

2 2 1, 1. � �x y z

20. Find the complete integral of the equation

2 2( ) �p q x pz

where , .z z

p qx y

� ��

� � (GATE-Maths, 1996)



21. Find the integral surface of the linear PDE

( ) ( )z z

x y x y z zx y

� ��

� � � �

which contains 1�z and 2 2 1. �x y (GATE-Maths, 1999)

Choose the correct answer in the following questions:

22. Using the transformation /u W y� in the PDE ,� x yxu u yu the transformed equationhas a solution of the form W =

(A) ( / )f x y (B) ( )f x y

(C) ( )�f x y (D) ( ).f xy (GATE-Maths, 1997)

23. The complete integral of the partial differential equation

3 2 2 3 3 3 2 2( ) 0 � �xp q yp q p q zp q

is z =

(A) 2 2( )� �� ax by ab ba (B) 2 2( )� �� ax by ab ba

(C) 2 2( )� �� ax by ba ab (D) 2 2( ).� �� ax by ab ba

(GATE-Maths, 1997)

24. The partial differential equation of the family of surfaces ( ) ( ) is� z x y A xy

(A) 0� �xp yq (B) � � �xp yq x y

(C) � � �xp yq x y (D) 0. �xp yq

(GATE-Maths, 1998)

25. The complete integral of the PDE sin ( )z px qy pq� � is

(A) sin ( )z ax by ab� (B) sin ( )z ax by ab� �

(C) sin ( )z ax y b� (D) sin ( ).z x by a� �(GATE-Maths, 2003)


��

Many practical problems in science and engineering, when formulated mathematically, giverise to partial differential equations (often referred to as PDE). In order to understand thephysical behaviour of the mathematical model, it is necessary to have some knowledge aboutthe mathematical character, properties, and the solution of the governing PDE. An equationwhich involves several independent variables (usually denoted by x, y, z, t, …), a dependentfunction u of these variables, and the partial derivatives of the dependent function u withrespect to the independent variables such as

( , , , , , , , , , , , , , , ) 0x y z t xx yy xyF x y z t u u u u u u u� � � � � (1.1)

is called a partial differential equation. A few well-known examples are:

(i) ( )t xx yy zzu k u u u� � � [linear three-dimensional heat equation]

(ii) 0xx yy zzu u u� � � [Laplace equation in three dimensions]

(iii) 2 ( )tt xx yy zzu c u u u� � � [linear three-dimensional wave equation]

(iv) μt x xxu uu u� � [nonlinear one-dimensional Burger equation].

In all these examples, u is the dependent function and the subscripts denote partial differentiationwith respect to these variables.

Definition 1.1 The order of the partial differential equation is the order of the highest derivativeoccurring in the equation. Thus the above examples are partial differential equations ofsecond order, whereas

sint xxxu uu x� �is an example for third order partial differential equation.

52

��

Fundamental Concepts


FUNDAMENTAL CONCEPTS 53

��

The most general linear second order PDE, with one dependent function u on a domain Ω of

points 1 2( , , , ), 1,nX x x x n� � � is

, 1 1

( )n n

ij xi j i xii j i

a u x b u F u G� �

� � ��

The classification of a PDE depends only on the highest order derivatives present.The classification of PDE is motivated by the classification of the quadratic equation of

the form

2 2 0Ax Bxy Cy Dx Ey F� � � � � � (1.3)

which is elliptic, parabolic, or hyperbolic according as the discriminant 2 4B AC� is negative,zero or positive. Thus, we have the following second order linear PDE in two variablesx and y:

xx xy yy x yAu Bu Cu Du Eu Fu G� � � � � � (1.4)

where the coefficients A, B, C, … may be functions of x and y, however, for the sake ofsimplicity we assume them to be constants. Equation (1.4) is elliptic, parabolic or hyperbolicat a point (x0, y0) according as the discriminant

20 0 0 0 0 0( , ) 4 ( , ) ( , )B x y A x y C x y�

is negative, zero or positive. If this is true at all points in a domain ,Ω then Eq. (1.4) is saidto be elliptic, parabolic or hyperbolic in that domain. If the number of independent variablesis two or three, a transformation can always be found to reduce the given PDE to a canonicalform (also called normal form). In general, when the number of independent variables isgreater than 3, it is not always possible to find such a transformation except in certain specialcases. The idea of reducing the given PDE to a canonical form is that the transformedequation assumes a simple form so that the subsequent analysis of solving the equation ismade easy.

��

Consider the most general transformation of the independent variables x and y of Eq. (1.4)

to new variables , ,ξ η where

( , ),ξ ξ x y� ( , )η η x y� (1.5)

such that the functions ξ and η are continuously differentiable and the Jacobian

( , )( ) 0

( , )

x yx y y x

x yJ

x y

��

� � � � �ξ ξξ η ξ η ξ ηη η

(1.6)



in the domain Ω where Eq. (1.4) holds. Using the chain rule of partial differentiation, thepartial derivatives become

2 2

2 2

2

( )

2

ξ η

ξ η

ξξ ξη ηη ξ η



ξ η

ξ η

ξ ξ η η ξ η

ξ ξ ξ η ξ η η η ξ η

ξ ξ η η ξ η

x x x

y y y

xx x x x x xx xx

xy x y x y y x x y xy xy

yy y y y y yy yy

u u u

u u u

u u u u u u

u u u u u u

u u u u u u

� �

� �

� � � � �

� � � � � �

� � � � � ��

Substituting these expressions into the original differential equation (1.4), we get

ξξ ξη ηη ξ ηAu Bu Cu Du Eu Fu G� � � � � � (1.8)

where

2 2

2 2

2 ( ) 2

,

ξ ξ ξ ξ

ξ η ξ η ξ η ξ η

η η η η

ξ ξ ξ ξ ξ

η η η η η

x x y y

x x x y y x y y

x x y y

xx xy yy x y

xx xy yy x y

A A B C

B A B C

C A B C

D A B C D E

E A B C D E

F F G G

� � �

� � � �

� � �

� � � � �

� � � � �

� � ��

It may be noted that the transformed equation (1.8) has the same form as that of the originalequation (1.4) under the general transformation (1.5).

Since the classification of Eq. (1.4) depends on the coefficients A, B and C, we can alsorewrite the equation in the form

( , , , , )xx xy yy x yAu Bu Cu H x y u u u� � � (1.10)

It can be shown easily that under the transformation (1.5), Eq. (1.10) takes one of thefollowing three canonical forms:

(i) ( , , , , )u u u u uξξ ηη ξ ηφ ξ η� � (1.11a)

or

, ( , , , , )u u u uξη ξ ηφ ξ η� in the hyperbolic case

(ii) ( , , , , )u u u u uξξ ηη ξ ηφ ξ η� � in the elliptic case (1.11b)

(iii) ( , , , , )u u u uξξ ξ ηφ ξ η� (1.11c)



or

( , , , , )u u u uηη ξ ηφ ξ η� in the parabolic case

We shall discuss in detail each of these cases separately.Using Eq. (1.9) it can also be verified that

2 2 24 ( ) ( 4 )x y y xB AC B ACξ η ξ η� � � �

and therefore we conclude that the transformation of the independent variables does notmodify the type of PDE.

�� !"�#�� $%�&��

Since the discriminant 2 4 0B AC� � for hyperbolic case, we set 0A � and 0C � in Eq. (1.9),

which will give us the coordinates ξ and η that reduce the given PDE to a canonical form

in which the coefficients of ,u uξξ ηη are zero. Thus we have

2 2

2 2

0

0

x x y y

x x y y

A A B C

C A B C

ξ ξ ξ ξ

η η η η

� � � �

� � � �

which, on rewriting, become

2

2

0

0

x x

y y

x x

y y

A B C

A B C

ξ ξξ ξ

η ηη η

� � � ��

� �

� � � ��

� �

Solving these equations for ( / )x yξ ξ and ( / ),x yη η we get

2

2

4

2

4

2

x

y

x

y

B B AC

A

B B AC

A

ξξ

ηη

� � ��

� � ��

The condition 2 4B AC� implies that the slopes of the curves 1 2( , ) , ( , )x y C x y Cξ η� � are

real. Thus, if 2 4 ,B AC� then at any point (x, y), there exists two real directions given by thetwo roots (1.12) along which the PDE (1.4) reduces to the canonical form. These are calledcharacteristic equations. Though there are two solutions for each quadratic, we have consideredonly one solution for each. Otherwise we will end up with the same two coordinates.



Along the curve 1( , ) ,x y cξ � we have

0x yd dx dy� � �ξ ξ ξHence,

x

y

dy

dx

ξξ� �

� � � ��

(1.13)

Similarly, along the curve 2( , ) ,x y cη � we have

x

y

dy

dx

ηη� �

� � � ��

(1.14)

Integrating Eqs. (1.13) and (1.14), we obtain the equations of family of characteristics 1( , )x y cξ �

and 2( , ) ,x y cη � which are called the characteristics of the PDE (1.4). Now to obtain the

canonical form for the given PDE, we substitute the expressions of ξ and η into Eq. (1.8)

which reduces to Eq. (1.11a).To make the ideas clearer, let us consider the following example:

3 10 3 0xx xy yyu u u� � �

Comparing with the standard PDE (1.4), we have 23, 10, 3, 4 64 0.A B C B AC� � � � � � Hence

the given equation is a hyperbolic PDE. The corresponding characteristics are:

2 4 1

2 3x

y

B B ACdy

dx A

ξξ

� ��

2 43

2x

y

B B ACdy

dx A

ηη

� ��

To find ξ and ,η we first solve for y by integrating the above equations. Thus, we get

13 ,y x c� � 21

3y x c� �

which give the constants as

1 3 ,c y x� � 2 /3c y x� �Therefore,

13 ,y x cξ � � � 21

3y x cη � � �



These are the characteristic lines for the given hyperbolic equation. In this example, thecharacteristics are found to be straight lines in the (x, y)-plane along which the initial data,impulses will propagate.

To find the canonical equation, we substitute the expressions for ξ and η into Eq. (1.9) to get

2 2 23( 3) 10( 3) (1) 3 0

2 ( ) 2

1 12(3) ( 3) 10 ( 3)(1) 1 2(3) (1) (1)

3 3

10 100 646 10 6 12

3 3 3

0, 0, 0, 0

x x y y

x x x y y x y y

A A B C

B A B C

C D E F

� � � �

� � � � � � � �

� � � � � � � � �

� � � �

� ��

� ��

� � � �

Hence, the required canonical form is

640

3uξη � or 0uξη �


( , ) ( ) ( )u f gξ η ξ η� �

where f and g are arbitrary. Going back to the original variables, the general solution is

( , ) ( 3 ) ( /3)u x y f y x g y x� � � �

��

For the parabolic equation, the discriminant 2 4 0,B AC� � which can be true

if 0B � and A or C is equal to zero. Suppose we set first 0A � in Eq. (1.9). Then weobtain

2 2 0x x y yA A B Cξ ξ ξ ξ� � � �

or2

0x x

y yA B C

ξ ξξ ξ� � � �

� � ��

which gives

2 4

2x

y

B B AC

A

� � ��

ξξ



Using the condition for parabolic case, we get

2x

y

BA

ξξ

= − (1.15)

Hence, to find the function ( , )x yξ ξ= which satisfies Eq. (1.15), we set

2x

y

dy Bdx A

ξξ

= − =

and get the implicit solution

1( , )x y Cξ =

In fact, one can verify that 0A = implies 0B = as follows:

2 ( ) 2x x x y y x y yB A B Cξ η ξ η ξ η ξ η= + + +

Since 2 4 0,B AC− = the above relation reduces to

2 2 ( ) 2

2( ) ( )

x x x y y x y y

x y x y

B A AC C

A C A C


ξ ξ η η

= + + +

= + +

However,

22 2

x

y

B AC CA A A

ξξ

= − = − = −

Hence,

2( ) ( ) 0x x x yB A A A Cξ ξ η η= − + =

We therefore choose ξ in such a way that both A and B are zero. Then η can be chosen in

any way we like as long as it is not parallel to the -ξ coordinate. In other words, we choose η such

that the Jacobian of the transformation is not zero. Thus we can write the canonical equation

for parabolic case by simply substituting ξ and η into Eq. (1.8) which reduces to either of

the forms (1.11c).To illustrate the procedure, we consider the following example:

2 22 xxx xy yyx u xyu y u e− + =

The discriminant 2 2 2 2 24 4 4 0,B AC x y x y− = − = and hence the given PDE is parabolic

everywhere. The characteristic equation is

22

2 2x

y

dy B xy ydx A xx

ξξ

= − = = − = −



On integration, we havexy c�

and hence xyξ � will satisfy the characteristic equation and we can choose .yη � To find

the canonical equation, we substitute the expressions for ξ and η into Eq. (1.9) to get

2 2 2 2 2 2 2 2

2

2 0

0, , 2

0, 0, x

A Ay Bxy cx x y x y y x

B C y D xy

E F G e

� � � � � � �

� � � �

� � �

Hence, the transformed equation is2 2 xy u xyu eηη ξ� �

or2 /2u u eξ η

ηη ξη ξ� �

The canonical form is, therefore,

/2 2

2 1u u eξ η

ηη ξξ

η η� �

�� !&�� $%�&��

Since the discriminant 2 4 0,B AC� � for elliptic case, the characteristic equations

2

2

4

2

4

2

B B ACdy

dx A

B B ACdy

dx A

� ��

� ��

give us complex conjugate coordinates, say ξ and .η Now, we make another transformation

from ( , )ξ η to ( , )α β so that

,2

ξ ηα ��2i

ξ ηβ ��

which give us the required canonical equation in the form (1.11b).To illustrate the procedure, we consider the following example:

2 0xx yyu x u� �

The discriminant 2 24 4 0.B AC x� � � � Hence, the given PDE is elliptic. The characteristicequations are



2 24 4

2 2

B B ACdy xix

dx A

� � ��

2 4

2

B B ACdyix

dx A

� ��

Integration of these equations yields

2

1,2

xiy c� �

2

22

xiy c � �

Hence, we may assume that

21,

2x iyξ � � 21

2x iyη � �

Now, introducing the second transformation

,2

ξ ηα ��2i

ξ ηβ ��

we obtain

2

,2

xα � yβ �

The canonical form can now be obtained by computing

2 2 2

2 2 2

2 ( ) 2 ( ) 0

1

0

0, 0

x x y y

x x x y y x y y

x x y y

xx xy yy x y

xx xy yy x y

A A c x

B A B c

C A B c x

D A B c D E

E A B c D E

F G

α βα α α

α β α β α β α β

β β β β

α α α α α

β β β β β

� � � �

� � � � �

� � � �

� � � � � �

� � � � � �

� �

Thus the required canonical equation is

2 2 0x u x u uαα ββ α� � �

or

2

uu u α

αα ββ α� � �



EXAMPLE 1.1 Classify and reduce the relation2 2

2 22xx xy yy x yy xy u xyu x u u ux y

− + = +

to a canonical form and solve it.

Solution The discriminant of the given PDE is2 2 2 2 24 4 4 0B AC x y x y− = − =

Hence the given equation is of a parabolic type. The characteristic equation is

22

2 2x

y

dy B xy xdx A yy

ξξ

−= − = = = −

Integration gives 2 21.x y c+ = Therefore, 2 2x yξ = + satisfies the characteristic equation. The

-η coordinate can be chosen arbitrarily so that it is not parallel to ,ξ i.e. the Jacobian of the

transformation is not zero. Thus we choose

2 2 ,x yξ = + 2yη =

To find the canonical equation, we compute2 2 2 2 2 2 2 2

2 2

4 8 4 0

0, 4 , 0

x x y yA A B C x y x y x y

B C x y D E F G

ξ ξ ξ ξ= + + = − + =

= = = = = =Hence, the required canonical equation is

2 24 0x y uηη = or 0uηη =

To solve this equation, we integrate it twice with respect to η to get

( ),u fη ξ= ( ) ( )u f gξ η ξ= +

where ( )f ξ and ( )g ξ are arbitrary functions of .ξ Now, going back to the original independent

variables, the required solution is2 2 2 2 2( ) ( )u y f x y g x y= + + +

EXAMPLE 1.2 Reduce the following equation to a canonical form:2 2(1 ) (1 ) 0xx yy x yx u y u xu yu+ + + + + =

Solution The discriminant of the given PDE is

2 2 24 4(1 ) (1 ) 0B AC x y− = − + + <



Hence the given PDE is an elliptic type. The characteristic equations are

2 22 2

2 2

2 2

2

4(1 ) (1 )4 12 2(1 ) 1

4 12 1

x ydy B B AC yidx A x x

dy B B AC yidx A x

− + +− − += = − = −+ +

+ − += =+


2 21

2 22

ln ( 1) ln ( 1)

ln ( 1) ln ( 1)

x x i y y c

x x i y y c

ξ

η

= + + − + + =

= + + + + + =

Introducing the second transformation

,2

ξ ηα +=2i

η ξβ −=

we obtain

2

2

ln ( 1)

ln ( 1)

x x

y y

α

β

= + +

= + +

Then the canonical form can be obtained by computing

2 2 1,x x y yA A B Cα α α α= + + = 0,B = 1,C = 0D E F G= = = =

Thus the canonical equation for the given PDE is

0u uαα ββ+ =

EXAMPLE 1.3 Reduce the following equation to a canonical form and hence solve it:22 sin cos cos 0xx xy yy yu xu xu xu− − − =

Solution Comparing with the general second order PDE (1.4), we have

1,

0,

A

D

=

=

2 sin ,

cos ,

B x

E x

= −

= −

2cos ,

0, 0

C x

F G

= −

= =

The discriminate 2 2 24 4 (sin cos ) 4 0.B AC x x− = + = > Hence the given PDE is hyperbolic.

The relevant characteristic equations are

2

2

4 sin 12

4 1 sin2

dy B B AC xdx A

dy B B AC xdx A

− −= = − −

+ −= = −




1cos ,y x x c� � � 2cosy x x c� � �

Thus, we choose the characteristic lines as

1cos ,x y x cξ � � � � 2cosx y x cη � � � � �In order to find the canonical equation, we compute

2 2

2 2

,

0

2 ( ) 2

2 (sin 1) (sin 1) 4 sin 2 cos 4

0 0, 0, 0, 0

x x y y

x x x y y x y y

A A B C

B A B C

x x x x

C D E F G

� � � �

� � � �

� � � � � � �

� � � � �

ξ ξ ξ ξ


Thus, the required canonical equation is

0uξη �

Integrating with respect to ,ξ we obtain

( )u fη η�

where f is arbitrary. Integrating once again with respect to ,η we have

( ) ( )u f d gη η ξ� ��or

( ) ( )u gψ η ξ� �

where ( )g ξ is another arbitrary function. Returning to the old variables x, y, the solution of

the given PDE is

( , ) ( cos ) ( cos )u x y y x x g y x xψ� � � � � �

EXAMPLE 1.4 Reduce the Tricomi equation

0,xx yyu xu� � 0x �

for all x, y to canonical form.

Solution The discriminant 2 4 4 .B AC x� � � Hence the given PDE is of mixed type:

hyperbolic for 0x � and elliptic for 0.x �

Case I In the half-plane 0,x � the characteristic equations are

2

2

4 2

2 2

4

2

x

y

x

y

dy B B AC xx

dx A

dy B B ACx

dx A

ξξ

ηη

� � � �

� � � �



Integration yields

3/21

3/22

2( )

3

2( )

3

y x c

y x c

� � �

� � � �

Therefore, the new coordinates are

31

32

3( , ) ( )

2

3( , ) ( )

2

x y y x c

x y y x c

ξ

η

� � � �

� � � �

which are cubic parabolas.In order to find the canonical equation, we compute

2 2

1/2

9 90 0

4 4

39 , 0, ( ) , 0

4

x x y yA A B C x x

B x C D x E F G

ξ ξ ξ ξ

�

� � � � � � � �

� � � � � � � � �

Thus, the required canonical equation is

1/2 1/23 39 ( ) ( ) 0

4 4xu x u x uξη ξ η

� ��

or

1( )

6 ( )u u uξη ξ ηξ η

� ��

Case II In the half-plane x > 0, the characteristic equations are given by

,dy

i xdx

� dyi x

dx� �

On integration, we have

33( , ) ( ) ,

2x y y i xξ � � 33

( , ) ( )2

x y y i xη � �

Introducing the second transformation

,2

ξ ηα ��2i

ξ ηβ ��

we obtain

3,

2yα � 3( )xβ � �



The corresponding normal or canonical form is

10

3u u uαα ββ ββ

� � �

EXAMPLE 1.5 Find the characteristics of the equation

22 sin ( ) 0xx xy yy yu u x u u� � � �

when it is of hyperbolic type.

Solution The discriminant B2 – 4AC = 4 – 4sin2x = 4cos2x. Hence for all x � (2n – 1)�/2,the given PDE is of hyperbolic type. The characteristic equations are

2 41 cos

2

dy B B ACx

dx A

��


1sin ,y x x c� � � 2siny x x c� � �

Thus, the characteristic equations are

sin ,y x xξ � � � siny x xη � � �

EXAMPLE 1.6 Reduce the following equation to a canonical form and hence solve it:

( ) 0xx xy yyyu x y u xu� � � �

Solution The discriminant2 2 24 ( ) 4 ( ) 0B AC x y xy x y� � � � � � �

Hence the given PDE is hyperbolic everywhere except along the line y = x; whereas on theline y = x, it is parabolic. When ,y x� the characteristic equations are

2 4 ( ) ( )

2 2

dy B B AC x y x y

dx A y

� � ��

Therefore,

1,dy

dx� dy x

dx y�


1,y x c� � 2 22y x c� �

Hence, the characteristic equations are

,y xξ � � 2 2y xη � �



These are straight lines and rectangular hyperbolas. The canonical form can be obtained bycomputing

2 2 20, 2 ( ) ,

0, 0, 2( ), 0

x x y yA A B C y x y x B x y

C D E x y F G

ξ ξ ξ ξ� � � � � � � � � � �

� � � � � �

Thus, the canonical equation for the given PDE is

22( ) 2( ) 0x y u x y uξη η� � � � �

or22 2( ) 0u uξη ηξ ξ� � � �

or

0u

u u� ��

� � �� ξη ηξ ξ

ξ ηIntegration yields

( )u

f��

�ξ ηη

Again integrating with respect to ,η we obtain

1( ) ( )u f d gη η ξ

ξ� ��

Hence,

2 2 2 21( ) ( ) ( )u f y x d y x g y x

y x� � � � �

� �is the general solution.

EXAMPLE 1.7 Classify and transform the following equation to a canonical form:

2 2sin ( ) sin (2 ) cos ( )xx xy yyx u x u x u x� � �

Solution The discriminant of the given PDE is

2 2 2 24 sin 2 4 sin cos 0B AC x x x� � � �

Hence, the given equation is of parabolic type. The characteristic equation is

cot2

dy Bx

dx A� �

Integration gives

1ln siny x c� �



Hence, the characteristic equations are:

ln sin ,y xξ � � yη �η is chosen in such a way that the Jacobian of the transformation is nonzero. Now thecanonical form can be obtained by computing

20, 0, cos , 1,

0, 0,

A B C x D

E F G x

� � � �

� � �Hence, the canonical equation is

2cos ( )x u u xηη ξ� �or

2( ) 1[1 ] sin ( )e u e uη ξ η ξηη ξ

� � ��

EXAMPLE 1.8 Show that the equation

2

2 1xx x tt

Nu u u

x a� �

where N and a are constants, is hyperbolic and obtain its canonical form.

Solution Comparing with the general PDE (1.4) and replacing y by t, we have 1, 0,A B� �21/ , 2 / ,C a D N x� � � and 0.E F G� � � The discriminant 2 24 4/ 0.B AC a� � � Hence, the

given PDE is hyperbolic. The characteristic equations are22 4/4 1

2 2

adt B B AC

dx A a

��

Therefore,

1,

dt

dx a� � 1dt

dx a�


1,x

t ca

� � � 2x

t ca

� �

Hence, the characteristic equations are

,x atξ � � x atη � �

The canonical form can be obtained by computing

22 0,

2 ( ) 2 4,

2 20, ,

x x t t

x x x t t x t t

x t x t

A A B C

B A B C

N NC D D E E D E

x x

ξ ξ ξ ξ


ξ ξ η η

� � � �

� � � � �

� � � � � � �



Thus, the canonical equation for the given PDE is

24 ( ) 0

Nu u u

xξη ξ η� � �

Expressing x in terms of ξ and ,η the required canonical equation is

( ) 0N

u u uξη ξ ηξ η� � �

�

EXAMPLE 1.9 Transform the following differential equation to a canonical form:

2 4 2 3 0xx xy yy x yu u u u u� � � � �

Solution The discriminant 2 4 12 0.B AC� � � � Hence, the given PDE is elliptic. Thecharacteristic equations are

2

2

41 3

2

41 3

2

dy B B ACi

dx A

dy B B ACi

dx A

� ��

� ��

Integration of these equations yields

1(1 3) ,y i x c� � � � 2(1 3)y i x c� � � �

Hence, we may take the characteristic equations in the form

(1 3) ,y i xξ � � � (1 3)y i xη � � �

In order to avoid calculations with complex variables, we introduce the second transformation

,2

ξ ηα ��2i

ξ ηβ ��

Therefore,

,y xα � � 3xβ �The canonical form can now be obtained by computing

2 2

2 2

3

2 ( ) 2 0

3

1

2 3

0, 0

x x y y

x x x y y x y y

x x y y

xx xy yy x y

xx xy yy x y

A A B C

B A B C

C A B C

D A B C D E

E A B C D E

F G

α α α α

α β α β α β α β

β β β β

α α α α α

β β β β β

� � � �

� � � � �

� � � �

� � � � � �

� � � � � �

� �



Thus the required canonical form is

3 3 2 3 0u u u uαα ββ α β� � � �or

1( 2 3 )

3u u u uαα ββ α β� � � �

��' �(��

Let

Lu φ� (1.16)

where L is a differential operator given by1

0 1 1( ) ( ) ( )

n n

nn n

d dL a x a x a x

dx dx

�

��

One way of introducing the adjoint differential operator L* associated with L is to form theproduct vLu and integrate it over the interval of interest. Let

[ ] *B BB

AA AvLu dx uL v dx� �� (1.17)

which is obtained after repeated integration by parts. Here, L* is the operator adjoint to L,where the functions u and v are completely arbitrary except that Lu and L*v should exist.

EXAMPLE 1.10 Let 2 2( ) ( / ) ( ) ( / ) ( ) ;Lu a x d u dx b x du dx c x u� � � construct its adjoint L*.

Solution Consider the equation2

2

2

2

( ) ( ) ( )

( ) ( ) ( )

B B

A A

B B B

A A A

d u duvLu dx v a x b x c x u dx

dxdx

d u duav dx bv dx cv u dx

dxdx

� ��

� �

� �

� �

� � �However,

2

2( ) ( ) ( )

[ ] ( )

[ ] [ ( ) ] ( )

( ) [ ( )] ( )

( ) ( )

� �

� ��

� � ��

� � �

�

� �

�

�

� �

� �

B B

A A

BBA

A

BB BA A

A

B BBA

A A

B B

A A

d u dav dx av u dx

dx dx

u va av u dx

u av u av u av dx

dubv dx u bv u bv dx

dx

cv u dx u cv dx



Therefore,

[ ( ) ( ) ( )] [( ) ( ) ( )]B BB

AA AvLu dx u av u av u bv u av bv cv dx� � � � � ��

Comparing this equation with Eq. (1.17), we get

* ( ) ( ) ( ) (2 ) ( )L v av bv cv av a b v a b c v� � � � � � � � ��

Therefore,2

2* (2 ) ( )

d dL a a b a b c

dxdx� � � � � ��

Consider the partial differential equation

( ) 2xx xy yy x yL u Au Bu Cu Du Eu Fu φ� � � � � � � (1.18)

which is valid in a region S of the xy-plane, where A, B, C, …, φ are functions of x and y.

In addition, linear boundary conditions of the general form

xu u fα β� �

are prescribed over the boundary curve S� of the region S. Let

[ ] *S S

vLu d uL v dσ σ� ��

where the integrated part [ ] is a line integral evaluated over ,S� the boundary of S, then L*is called the adjoint operator. In general, a second order linear partial differential operator Lis denoted by

2

, 1 1

( )n n

ij ii j ii j i

u uL u A B Cu

x x x

� ��

� � �� (1.19)

Its adjoint operator is defined by

2

, 1 1

* ( ) ( ) ( )n n

ij ii j ii j i

L v A v B v Cvx x x� �

� � ��

(1.20)

Here it is assumed that (2)ijA C� and (1).iB C� For any pair of functions (2), ,u v C� it can

be shown that

1 1 1

( ) * ( )� � �

� ��

� � �n n n

ijij i

i j j ji j j

Au vvL u uL v A v u uv B

x x x x

�� (1.21)

This is known as Lagrange’s identity.



EXAMPLE 1.11 Construct an adjoint to the Laplace operator given by

( ) xx yyL u u u� � (1.22)

Solution Comparing Eq. (1.22) with the general linear PDE (1.19), we have

11 221, 1.A A� � From Eq. (1.20), the adjoint of (1.22) is given by

2 2

2 2* ( ) ( ) ( ) xx yyL v v v v v

x y� � � ��

Therefore,

* ( ) xx yyL u u u� �

Hence, the Laplace operator is a self-adjoint operator.

EXAMPLE 1.12 Find the adjoint of the differential operator

( ) xx tL u u u� � (1.23)

Solution Comparing Eq. (1.23) with the general second order PDE (1.19), we have

11 1,A � 1 1.B � � From Eq. (1.20), the adjoint of (1.23) is given by

2

2* ( ) ( ) ( ) xx tL v v v v v

tx� � � � �

� ��

Therefore,

* ( ) xx tL u u u� �It may be noted that the diffusion operator is not a self-adjoint operator.

��

In Section 1.2, we have noted with interest that a linear second order PDE

( ) ( , )L u G x y�

is classified as hyperbolic if 2 4 ,B AC� and it has two families of real characteristic curvesin the xy-plane whose equations are

1 1( , ) ,f x y cξ � � 2 2( , )f x y cη � �

Here, ( , )ξ η are the natural coordinates for the hyperbolic system. In the xy-plane, the curves

1( , )x y cξ � and 2( , )x y cη � are the characteristics of the given PDE as shown in Fig. 1.1(a),

while in the -plane,ξη the curves 1cξ � and 2cη � are families of straight lines parallel to

the axes as shown in Fig. 1.1(b).



O

Characteristics

x

y

� = const.

� = const. �

�

Characteristics

O

(a) (b)Fig. 1.1 Families of characteristic lines.

A linear second order partial differential equation in two variables, once classified as ahyperbolic equation, can always be reduced to the canonical form

2

( , , , , )x yu

F x y u u ux y

��

�

In particular, consider an equation which is already reduced to its canonical form in thevariables x, y:

2

( ) ( , )u u u

L u a b cu F x yx y x y

� � ��

� � � � � (1.24)

where L is a linear differential operator and a, b, c, F are functions of x and y only and are

differentiable in some domain IR .

Let ( , )v x y be an arbitrary function having continuous second order partial derivatives.Let us consider the adjoint operator L* of L defined by

2

* ( ) ( ) ( )v

L v av bv cvx y x y

� � ��

� � � � (1.25)

Now we introduce

,v

M auv uy

��

� �u

N buv vx

��

� � (1.26)

then

( ) ( ) ( ) ( )x y x x x y xy y y y x xyM N u av u av u v uv u bv u bv v u vu� � � � � � � � �



Adding and subtracting cuv, we get

2 2

( ) ( )x yv u u u

M N u av bv cv v a b cux y x y x y x y

� � � ��

� � � �

� � � � � ��

i.e.

* x yvLu uL v M N� � � (1.27)

This is known as Lagrange identity which will be used in the subsequent discussion. Theoperator L is a self-adjoint if and only if L = L*. Now we shall attempt to solve Cauchy’sproblem which is described as follows: Let

( ) ( , )L u F x y� (1.28)

with the condition (Cauchy data)

(i) ( )u f x� on ,Γ a curve in the xy-plane;

(ii) ( )u

g xn��

� on .�

That is u, and its normal derivatives are prescribed on a curve � which is not a characteristic line.

Let � be a smooth initial curve which is also continuous as shown in Fig. 1.2. SinceEq. (1.24) is in canonical form, x and y are the characteristic coordinates. We also assume that

the tangent to � is nowhere parallel to the coordinate axes.

Ox

y

IR

R

P( , )��

Data curve

Q

Fig. 1.2 Cauchy data.

Let ( , )P ξ η be a point at which the solution to the Cauchy problem is sought. Let us draw

the characteristics PQ and PR through P to meet the curve Γ at Q and R. We assume that

u, ux, uy are prescribed along .Γ Let IR� be a closed contour PQRP bounding IR. Since

Eq. (1.28) is already in canonical form, the characteristics are lines parallel to x and y axes.Using Green’s theorem, we have

IR( ) ( )x y

R

M N dxdy M dy N dx�

� � �� (1.29)



where IR∂ is the boundary of IR . Applying this theorem to the surface integral of Eq. (1.27),

we obtain

IRIR

( ) [ ( ) * ( )]M dy N dx vL u uL v dx dy− = −∫ ∫∫∂(1.30)

In other words,

( ) ( ) ( ) [ ( ) * ( )]- + - + - = -Ú Ú Ú ÚÚRP PQR

M dy N dx M dy N dx M dy N dx vL u uL v dx dyΓ

Now using the fact that 0dy = on PQ and 0dx = on PR, we have

IR

( ) [ ( ) * ( )]RP PQ

M dy N dx M dy N dx vL u uL v dx dyΓ

− + − = −∫ ∫ ∫ ∫∫ (1.31)

From Eq. (1.26), we find thatQ Q

xPQ P PN dx buvdx vu dx= +∫ ∫ ∫

Integrating by parts the second term on the right-hand side and grouping, the above equationbecomes

[ ] ( )= + -Ú ÚQ

Qp x

PQ PN dx uv u bv v dx

Substituting this result into Eq. (1.31), we obtain

IR

[ ] [ ] ( ) ( )

( ) [ ( ) * ( )]

p Q x yPQ RPuv uv u bv v dx u av v dy

M dy N dx vL u uL v dxdyΓ

= + − − −

− − + −

∫ ∫

∫ ∫∫ (1.32)

Let us choose ( , ; , )v x y ξ η to be a solution of the adjoint equation

*( ) 0L v = (1.33)

and at the same time satisfy the following conditions:

(i) xv bv= when ,y η= i.e., on PQ (1.34a)

(ii) yv av= when ,x ξ= i.e., on PR (1.34b)

(iii) 1v = when ,x ξ= y η= (1.34c)

We call this function ( , ; , )v x y ξ η as the Riemann function or the Riemann-Green function.

Since ( ) ,L u F= Eq. (1.32) reduces to



IR

[ ] [ ] [ ( ) ( ) ] ( )� � � � � �P Q y xu uv u av v dy v bu u dx vF dx dy�� (1.35)

This is called the Riemann-Green solution for the Cauchy problem described by Eq. (1.28)

when u and ux are prescribed on .� Equation (1.35) can also be written as

IR

[ ] [ ] ( ) ( ) ( )� �

� � � � � �� P Q y xu uv uv a dy b dx uv dy vu dx vF dx dy (1.36)

This relation gives us the value of u at a point P when u and ux are prescribed on .Γ But whenu and uy are prescribed on ,Γ we obtain

IR

[ ] [ ] ( ) ( ) ( )P R x yu uv uv a dy b dx uv dx vu dy vF dx dyΓ Γ

� � � � � �� (1.37)

By adding Eqs. (1.36) and (1.37), the value of u at P is given by

IR

1 1[ ] {[ ] [ ] } ( ) ( )

2 2

1( ) ( )

2

P Q R x y

x y

u uv uv uv a dy b dx u v dx v dy

v u dx u dy vF dx dy

� �

�

� � � � � �

� � �

� �

� �� (1.38)

Thus, we can see that the solution to the Cauchy problem at a point ( , )ξ η depends only

on the Cauchy data on .Γ The knowledge of the Riemann-Green function therefore enablesus to solve Eq. (1.28) with the Cauchy data prescribed on a noncharacteristic curve.

EXAMPLE 1.13 Obtain the Riemann solution for the equation2

( , )u

F x yx y

��

�

given

(i) ( )u f x� on Γ

(ii) ( )u

g xn

��

� on Γ

where Γ is the curve .y x�

Solution Here, the given PDE is

2

( ) ( , )u

L u F x yx y

��

� � (1.39)



We construct the adjoint L* of L as follows: setting

,yM auv uv� � xN buv vu� �

and comparing the given equation (1.39) with the standard canonical form of hyperbolicequation (1.24), we have

0a b c� � �

Therefore,

,yM uv� � xN vu� (1.40)

and

( ) * ( )x y xy xyM N vu uv vL u uL v� � � � �

Thus,

2

* ( )v

L vx y

��

� (1.41)

Here, L = L* and is a self-adjoint operator. Using Green’s theorem

IRIR

( ) ( )x yM N dx dy M dy N dx�

� � �� we have

IRIR

[ ( ) * ( )] ( )vL u uL v dx dy M dy N dx�

� � �� or

IRIR

[ * ( )] ( )vF uL v dx dy M dy N dx�

� � �� (1.42)

But

IR( ) ( ) ( ) ( )

QP PRM dy N dx M dy N dx M dy N dx M dy N dx

��

(1.43)

where

( )v u

M dy N dx u dy v dxy xΓ Γ

� � � ��

� �

From Fig. 1.3, we have on ,Γ .x y� Therefore, .dx dy� Hence

( )v u

M dy N dx u v dxy xΓ Γ

� � � ��

� �(1.44)



Ox

y

R

P , ( )ξ η Q

x = ξy =

x

y = η

Fig. 1.3 An illustration of Example 1.13.

since on QP, y = constant. Therefore, 0.dy = Thus,

( )QP QP QP

uM dy N dx N dx v dxx∂− = − = −∂∫ ∫ ∫ (1.45)

Similarly, on PR, x = constant. Hence, 0.dx = Thus,

( )PR PR PR

vM dy N dx M dy u dyy

− = = −∫ ∫ ∫ ∂∂

(1.46)

Substituting Eqs. (1.44)–(1.46) into Eq. (1.43), we obtain from Eq. (1.42), the relation

IR

[ * ( )]QP PR

v u u vvF uL v dx dy u dx v dx v dx u dyy x x y

⎛ ⎞− = − − + − + −⎜ ⎟⎝ ⎠∫∫ ∫ ∫ ∫Γ

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

But

[ ] PQQP QP

u vv dx vu u dxx x

∂ ∂∂ ∂

− = − +∫ ∫Therefore,

IR

[ * ( )]

[ ] PQ QP PR

v uvF uL v dx dy u dx v dxy x

v vvu u dx u dyx y

⎛ ⎞− = − −⎜ ⎟⎝ ⎠

+ − + + −

∫∫ ∫

∫ ∫

Γ

∂ ∂∂ ∂

∂ ∂∂ ∂ (1.47)

Now choosing ( , ; , )v x y ξ η as the solution of the adjoint equation such that

(i) * 0L v = throughout the xy-plane

(ii) 0vx=∂

∂ when ,y η= i.e., on QP



(iii) 0v

y

��

� when ,x ξ� i.e., on PR

(iv) 1v � at ( , ).P ξ η

Equation (1.47) becomes

IR

( ) ( ) ( )Q Pv u

vF dx dy u dx v dx uv uy x�

� ��

� �

or

IR

( ) ( ) ( )P Qv u

u uv u dx v dx v F dx dyy x�

� ��

� � (1.48)

However,

( ) ( ) ( ) ( ) ( )

( )

Q R

x x y y

uv uv d uv uv dx uv dyx y

u v dx uv dx u v dy uv dy

� �

�

� ��

�

� � � �

� �

�

� ��

Now Eq. (1.48) can be rewritten as

IR

( ) ( ) ( ) ( )P R y yu uv u v dy uv dy vF dxdy�

� � � �� (1.49)

Finally, adding Eqs. (1.48) and (1.49), we get

IR

1 1( ) [( ) ( ) ] ( )

2 2

1( ) ( )

2

P Q R x x

y y

u uv uv uv dx vu dx

u v dy uv dy vF dx dy

�

�

� � � � �

� � �

�

� ��

EXAMPLE 1.14 Verify that the Green function for the equation

2 20

u u u

x y x y x y

� ��

� � ��

subject to 20, / 3u u x x� �� on ,y x� is given by

3

( ) {2 ( ) ( ) 2 }( , ; , )

( )

x y xy x yv x y

ξ η ξηξ ηξ η

� � � � ��

�

and obtain the solution of the equation in the form2 2( ) (2 2 )u x y x xy y� � � �



Solution In the given problem,

2 2 2( ) 0

u u uL u

x y x y x x y y� � � �

� ��

(1.50)

Comparing this equation with the standard canonical hyperbolic equation (1.24), we have

2,a b

x y� �

�0,C � 0F �

Its adjoint equation is * ( ) 0,L v � where

2 2 2* ( ) .

v v vL v

x y x x y y x y

� � ��

� � � ��

(1.51)

such that

(i) * 0L v � throughout the xy-plane

(ii)2v

vx x y

��

��

on PQ, i.e., on y η�

(iii)2v

vy x y

��

��

on PR, i.e., on x ξ�

(1.52)

(iv) 1v � at ( , ).P ξ η

If v is defined by

3

( )( , ; , ) [2 ( ) ( ) 2 ]

( )

x yv x y xy x yξ η ξ η ξη

ξ η��

(1.53)

Then

3 3

2 ( ) ( ) 2[2 ( )]

( ) ( )

v x y xy x yy

x

� � ��

� � � �

� ��

��

or

23

1[4 2 2 ( ) 2 ]

( )

vxy y x

x� � ��

� ��

��

(1.54)

and

2

3

23

4( )

( )

1[4 2 2 ( ) 2 ]

( )

v x y

x y

vxy x y

y

��

��

��

� � � � ��

ξ η

ξ η ξηξ η

(1.55)

(1.56)



Using the results described by Eqs. (1.53)–(1.56), Eq. (1.51) becomes

2

2

2 23 3

2 4* ( )

( )

4( ) 2[4 2( )]

( ) ( ) ( )

v v v vL v

x y x y x y x y

x yxy x y

x yξ η ξ η

� � � ��

��

� � ��

or

3 3

4( ) 4( )* ( ) 0

( ) ( )

x y x yL v

� � � �

� ��

� �

Hence condition (i) of Eq. (1.52) is satisfied. Also, on .y η�

23

1[4 2 2 ( ) 2 ]

( )

vx x

x� � � � ��

� ��

��

or

23

1[2 2 ( ) 2 ]

( )y

vx

x

��

� � � ��η

η ξ η ξηξ η

��

Also, 2 /( )v x y� at y η� is given by

3

23

2 2[2 ( ) ( ) 2 ]

( )

1[2 2 ( ) 2 ]

( )

v xx x

x y x

x

η η ξ η η ξηη ξ η

η ξ η ξηξ η

��

� � � ��

(1.58)

From Eqs. (1.57) and (1.58), we get

2vv

x x y�

��

at y η�

Thus, property (ii) in Eq. (1.52) has been verified. Similarly, property (iii) can also be

verified. Also, at , ,x yξ η� �

22

3 3

( ) ( )[2 ( ) 2 ] 1

( ) ( )v

ξ η ξ η ξ ηξη ξ η ξηξ η ξ η� � ��

Thus property (iv) in Eq. (1.52) has also been verified.



From Eqs. (1.50) and (1.51), we have2 2 2 2

( ) * ( )

2 2

2 2

u v vu vuvL u uL v v u

x y x y x x y y x y

u v vu vuv u

y x x y x x y y x y

vu v vu uu v

x x y y y x y x

M N

x y

� � � ��

� � � � � ��

� � � ��

� �

� � � ��

� � � � � ��

� � � ��

� ��

where

2,

uv vM u

x y y� �

��

2vu uN v

x y x� �

��

Now using Green’s theorem, we have

IRIR

[ ( ) * ( )] ( ) ( )

( ) ( )

Q

R

P R

Q P

vL u uL v dx dy M dy N dx M dy N dx

M dy N dx M dy N dx

� � � � �

� � � �

��

� �

�

��

(see Fig. 1.3) on QP, .y C� Hence, 0;dy � on PR, .x C� Therefore, 0dx �

2 2

2 2

Q

R

P R

Q P

uv v uv uu dy v dx

x y y x y x

uv u uv vv dx u dy

x y x x y y

� ��

� � � ��

�

� �

� ��

� ��

However,

2 2( )

P P PPQQ Q Q

uv u uv vv dx dx uv u dx

x y x x y x

� ��

� � � ��

Now, using the condition 0u � on ,y x� Eq. (1.59) becomes

IR

2 2[ ( ) * ( )]

2( ) ( )

2

Q Q

R R

P P

P QQ Q

P R

Q P

uv v uv uvL u uL v dx dy u dy v dx

x y y x y x

uv vdx uv uv u dx

x y x

uv vdy u dy

x y y

� � � ��

� � � ��

� ��

��

� �

� �

� ��

��

��



Also, using conditions (ii)–(iv) of Eq. (1.52), the above equation simplifies to

( ) ( )Q

P QR

uu uv v dx

x� � � �

�Now using the given condition, viz.

23 onu

x RQx

��

�

we obtain

22

3

5 33

6 6 4 43

2 24 2 2 4 2 2

3

2 2

2 [2 2 ]( ) ( ) 3

( )

12( )

( )

12 1 1( ) ( )

6 4( )

[2( ) 3 ( )]( )

( ) (2 2 )

Q

P QR

x xu uv x dx

x x dx�

�

��

� �

��

� � ��

� ��

� �

� � � ��

� ��

� � ��

� ��

��

� � � �

�

�

Therefore,2 2( , ) ( ) (2 2 )u x y x y x xy y� � � �

Hence the result.

EXAMPLE 1.15 Show that the Green’s function for the equation2

0u

ux y

��

� �

is

0( , ; , ) 2( ) ( )v x y J x y� � � ��

where J0 denotes Bessel’s function of the first kind of order zero.

Solution Comparing with the standard canonical hyperbolic equation (1.24), we have

0,a b� � 1c �

It is a self-adjoint equation and, therefore, the Green’s function v can be obtained from

2

0v

vx y

��

� �



subject to

0 on

0 on

1 at ,

vy

x

vx

y

v x y

��

��

� �

� �

� � �

η

ξ

ξ ηLet

( ) ( )k a x y

v v

x x

� � ��

� � �

�

φ ξ η

φφ

But

1 ( )kk a yx

��

� � �φφ η

Therefore,

1 ( )kay

x k

��

�� φ φ η

Thus,

1

21

21 1

2

( )

( )

(1 ) ( ) ( )

k

k

k k k

v v ay

x k

v v ay

x y y k

a v v vk y y

k y y

� ��

� � ��

� � � � ��

�

�

� � �

� �

� ��

� �

� ��

� �

φ ηφ

φ ηφ

φ φφ φ η φ ηφ φ φ

However,

1 ( )kax

y k

��

�� φ φ ξ

Therefore,

2 21 1 1 1

2(1 ) ( ) ( ) ( ) ( )k k k k kv a v a v v a

k x y y xx y k k k

� � � ��

� � � � ��

� �φ φ ξ η φ φ η φ ξ

φ φ φ



Hence,2

0v vx y

+ =∂∂ ∂

gives

2 12(1 ) 1

2 (1 ) 0k k

k ka d v dv dvk vk k k d dd

φ φφ φφ φφ

−− −⎡ ⎤

+ − + + =⎢ ⎥⎢ ⎥⎣ ⎦

or

22 1

2 2 0k ka d v dv vdk d

φ φφφ

− −⎛ ⎞+ + =⎜ ⎟⎝ ⎠

or

22 0kkv v v

aφ φ φ+ + =′′ ′

Let 2, 4.k a= = Then the above equation reduces to

2 2 10 (Bessel’s equation)v v v v v vφ φ φφ

+ + = = + +′′ ′ ′′ ′

Its solution is known to be of the form

0 0( ) 2( )( )v J J x yφ ξ η= = − −

which is the desired Green’s function.

1.6 LINEAR PARTIAL DIFFERENTIAL EQUATIONS WITH CONSTANTCOEFFICIENTS

An nth order linear PDE with constant coefficients can be written in the form

0 1 21 2 2

n n n n

nn n n n

u u u ua a a a

x x y x y y- -∂ ∂ ∂ ∂+ + + +∂ ∂ ∂ ∂ ∂ ∂

L

= f(x, y) (1.60)

where a0, a1, …, an are constants; u is the dependent variable; x and y are independent

variables. Introducing the standard differential operator notation, such as Dx

∂=∂

, D¢ = y

∂∂

,

the above equation can be rewritten as

F(D, D¢)u = (a0Dn + a1D

n–1D¢ + a2Dn–2D¢2 + … + anD¢n)u = f(x, y) (1.61)



It can also be written in more compact form as

( , ) i jij

i j

F D D u C D D=¢ ¢Â Â (1.62)

where Cij are constants, , , ,n n

n nn n

D D D Dx y x y

∂ ∂ ∂ ∂= = = =¢ ¢∂ ∂ ∂ ∂

, etc.

As in the case of linear ODE with constant coefficients, the complete solution of Eq. (1.61)consists of two parts:

(i) the complementary function (CF), which is the most general solution of theequation F(D, D¢)u = 0, the one containing, n arbitrary functions, where n is theorder of the DE.

(ii) the particular integral (PI), is a particular solution, which is free from arbitraryconstants or functions of the equation F(D, D¢)u = f(x, y).

The complete solution of Eq. (1.61) is then

u = CF + PI (1.63)

It may be noted that, if all the terms on the left-hand side of Eq. (1.61) are of the sameorder, it is said to be a homogeneous equation otherwise, it is a non-homogeneous equation.Now, we shall study few basic theorems as is the case in ODEs.

Theorem 1.1 If uCF and uPI are respectively the complementary function and particularintegral of a linear PDE, then their sum (uCF + uPI) is a general solution of the given PDE.

Proof Since F(D, D¢)uCF = 0,

and F(D, D¢)uPI = f(x, y),

we arrive at F(D, D¢)uCF + F(D, D¢)uPI = f(x, y).

showing that (uCF + uPI) is in fact a general solution of Eq. (1.61). Hence proved.

Theorem 1.2 If u1, u2, …, un are the solutions of the homogeneous PDE: F(D, D¢)u = 0,

then 1

n

i ii

C u=Â , where Ci are arbitrary constants, is also a solution.

Proof Since we observe that

F(D, D¢)(Ciui) = CiF(D, D¢)ui

and1 1

( , ) ( , )n n

i ii i

F D D v F D D v= =

=¢ ¢Â Â



For any set of functions vi, we find at once,

1 1

( , ) ( , )( )n n

i i i ii i

F D D C u F D D C u� �

��

1

( , ) 0n

i ii

C F D D u�

� ��Hence proved.

We shall now classify linear differential operator F(D, D�) into reducible and irreducibletypes, in the sense that F(D, D�) is reducible if it can be expressed as the product of linearfactors of the form (aD + bD� + c), where a, b and c are constants, otherwise F(D, D�) isirreducible. For example, the operator

F(D, D�)u = (D2 – D�2 + 3D + 2D� + 2)u

= (D + D� + 1)(D – D� + 2)u

is reducible. While the operator F(D, D�)u = (D2 – D�), is irreducible, due to the fact that itcannot be factored into linear factors.

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The general strategy adopted for finding the CF of reducible equations is stated in thefollowing theorems:

Theorem 1.3 If the operator F(D, D�) is reducible that is, if (aiD + biD� + ci) is a factor ofF(D, D�) and �i(�) is an arbitrary function of a single variable �, then, if ai 0,

ui = exp i

i

cx

a

� ��

� �i(bix – aiy) (1.64)

is a solution of the equation F(D, D�)u = 0 (Sneddon, 1986).

Proof Using product rule of differentiation, Eq. (1.64) gives

exp ( )i ii i i i

i i

c cDu x b x a y

a a�

� � � � � � ��

exp ( )ii i i i

i

cb x b x a y

a�

� � � ��

exp ( )i ii i i i

i i

c cu b x b x a y

a a�

� ��

.

Similarly, we get

exp ( )ii i i i

i

cD u a x b x a y

a�

� ��

.



Thus, we observe that

(aiD + biD¢ + ci)ui = 0 (1.65)

That is, if the operator F(D, D¢) is reducible, the order in which the linear factors appearis immaterial. Thus, if

1

( , ) ( ) ( )n

i j j j i i i ij

F D D u a D b D c a D b D c u=

Ï ¸Ô Ô¢= + + + +¢ ¢ ¢Ì ˝Ô ÔÓ ˛’ (1.66)

where, the prime on the product indicates that the factor corresponding to i = j is omitted.Combining Eqs. (1.65) and (1.66), we arrive at the result F(D, D¢)ui = 0. Hence proved.

It may be noted that if no two factors of Eq. (1.65) are linearly independent, then thegeneral solution of Eq. (1.66) is the sum of the general solutions of the equations of the form(1.65). For illustration, we consider the following examples:

EXAMPLE 1.16 Solve the following equation (D2 + 2DD¢ + D¢2 – 2D – 2D¢)u = 0.

Solution Observe that the given PDE is non-homogeneous and can be factored as

(D2 + 2DD¢ + D¢2 – 2D – 2D¢)u = (D + D¢)(D + D¢ – 2)u.

Using the result of Theorem 1.3, we get the general solution or the CF as

uCF = f1(x – y) + e2xf2(x – y).

On similar lines, we can also establish the following result:

Theorem 1.4 (Sneddon, 1986) Let (bi D¢ + ci) is a factor of F(D, D¢)u, and fi(x) is anarbitrary function of a single variable x, then, if bi π 0, we have

ui = exp i

i

cy

b

Ê ˆ-Á ˜Ë ¯

fi(bix) (1.67)

as a solution of the equation F(D, D¢)u = 0.Proof Suppose, the factorisation of F(D, D¢) = 0 gives rise to a multiple factors of the

form (aiD + biD¢ + ci)n, the solution of F(D, D¢)u = 0 can be obtained by the application of

Theorems 1.3 and 1.4. For example, let us find the solution of

(aiD + biD¢ + ci)2u = 0 (1.68)

We set,

(aiD + biD¢ + ci)u = U,

then, Eq. (1.68) becomes

(aiD + biD¢ + ci)U = 0.

Using Theorem 1.3, its solution is found to be

U = exp i

i

cx

a

Ê ˆ-Á ˜Ë ¯

fi(bix – aiy).



Further, assume that ai π 0, now, in order to find u, we have to solve

(aiD + biD¢ + ci)u = exp i

i

cx

a

Ê ˆ-Á ˜Ë ¯

fi(bix – aiy).

This is a first order PDE. Using the Lagrange’s method (Section 0.8), its auxiliary equationsare

exp ( )i i ii i i i

i

dx dy du

a b cc u x b x a y

af

= =Ê ˆ

- + - -Á ˜Ë ¯

(1.69)

One solution of which is given by

bix – aiy = l (a constant) (1.70)

Substituting this solution into the first and third of the auxiliary equations, we obtain

exp ( )i ii i

i

dx du

a cc u x

af l

=Ê ˆ

- + -Á ˜Ë ¯

or 1

exp ( )i ii

i i i

c cduu x

dx a a af l

Ê ˆ+ = -Á ˜Ë ¯

.

This being an ODE, its solution can be readily written as

1exp ( ) (constant)i

ii i

cu x x

a af l m

Ê ˆ= +Á ˜Ë ¯

or 1

[ ( ) ] exp ii

i i

cu x x

a af l m

Ê ˆ= + -Á ˜Ë ¯

(1.71)

Thus, the solution of Eq. (1.68) is given by

u = [xfi(bix – aiy) + yi(bix – aiy]e–ci/aix (1.72)

where fi and yi are arbitrary functions.In general, if there are n, multiple factors of the form (aiD + biD¢ + ci), then the solution

of (aiD + biD¢ + ci)n u = 0 can be written as

1

1

exp ( )n

jiij i i

i j

cu x x b x a y

af-

=

È ˘Ê ˆÍ ˙= - -Á ˜Ë ¯ Í ˙Î ˚Â (1.73)

Here follows an example for illustration.



EXAMPLE 1.17 Find the solution of the equation (2D – D� + 4)(D + 2D� + 1)2u = 0.

Solution The complementary function (CF) corresponding to the factor (2D – D� + 4)is e–2x �(–x – 2y). Similarly, CF corresponding to

(D + 2D� + 1)2 is e–x[�1(2x – y) + x�2(2x – y)].

Thus, the CF for the given PDE is given by

u = e–2x�(x + 2y) + e–x[�1(2x – y) + x�2(2x – y)],

where, �, �1, �2 are arbitrary functions.

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If the operator F(D, D�) is irreducible, we can find the complementary function, containingas many arbitrary functions as we wish by a method which is stated in the following theorem:

Theorem 1.5 The solution of irreducible PDE F(D, D�)u = 0 is

1

exp( )i i ii

u c a x b y�

�

� �� (1.74)

Proof Let us assume the solution of F(D, D�)u = 0 in the form, u = ceax+by, where a, b andc are constants to be determined. Then, we have

Diu = caieax+by, D�ju = cbj eax+by,

DiD�ju = caib jeax+by

Thus, F(D, D�)u = 0 yields c[F(a, b)]eax+by = 0

where c is an arbitrary constant, not zero, holds true iff

F(a, b) = 0, (1.75)

indicating that there exists infinite pair of values (ai, bi) satisfying Eq. (1.75). Hence,

1

i ia x b yi

i

u c e�

�

�

�� (1.76)

is a solution of irreducible equationF(D, D�)u = 0, (1.77)

provided F(ai, bi) = 0 (1.78)

It may be noted that this method is applicable even for reducible equations. Here followsan example for illustration:

EXAMPLE 1.18 Solve the following equation (2D2 – D�2 + D)u = 0.

Solution The given equation is an irreducible non-homogeneous PDE. Using the resultof Theorem 1.5, it follows immediately that



CF1

i ia x b yi

i

u u c e•

+

=

= =Âwhere ai, bi are related by F(ai, bi) = 0.

That is,2ai

2 – bi2 + ai = 0

which gives bi2 = 2ai

2 + ai.

1.6.3 Methods for Finding the Particular Integral (PI)

To find the PI of Eq. (1.61), we rewrite the same in the form

1( , )

( , )u f x y

F D D=

¢(1.79)

Very often, the operator F–1(D, D¢) can be expanded, using binomial theorem and interpretthe operators D–1, (D¢)–1 as integrations. That is,

1

constant

1( , ) ( , ) ( , )

y

D f x y f x y f x y dxD

- = = Ú ,

andconstant

1( , ) ( , )

x

f x y f x y dyD

=¢ Ú .

We present below different cases for finding the PI, depending on the nature of f(x, y).

Case I Let f(x, y) = exp(ax + by), then

1 1

( , ) ( , )ax by ax bye e

F D D F a b+ +=

¢(1.80)

By direct differentiation, we find DiD¢jeax+by = aib j eax+by.In other words,

F(D, D¢)eax+by = F(a, b)eax+by,that is,

1( , )

( , )ax by ax bye F a b e

F D D+ +=

¢.

Dividing both sides by F(a, b), we get

1 1

( , ) ( , )ax by ax bye e

F D D F a b+ +=

¢ ,

provided F(a, b) π 0.

Case II Let f(x, y) = sin(ax + by) or cos(ax + by), where a and b are constants, then, since

D2 sin(ax + by) = –a2 sin(ax + by)

DD¢ sin(ax + by) = –ab sin(ax + by)



D¢2 sin(ax + by) = –b2 sin(ax + by)

We notice that

1

( , )F D D¢ sin(ax + by)

is obtained by setting, D2 = –a2, DD¢ = –ab, D¢2 = –b2 provided F(D, D¢) π 0. Thus,

F(D2, DD¢, D¢2) sin(ax + by) = F(–a2, –ab, –b2) sin(ax + by)

or2 2 2 2

1 1sin( ) sin( )

( , , ) ( , , )ax by ax by

F D DD D F a ab b+ = +

- - -¢ ¢(1.81)

Similarly,

2 2 2 2

1 1cos( ) cos( )

( , , ) ( , , )ax by ax by

F D DD D F a ab b+ = +

- - -¢ ¢(1.82)

Case III Let f(x, y) is of the form xpyq, where p and q are positive integers. Then, the PIcan be obtained by expanding F(D, D¢) in ascending powers of D or D¢.Case IV Let f(x, y) is of the form eax+by f(x, y).

Then,F(D, D¢)[eax+by f(x, y)] = eax+by F(D + a, D¢ + b) f(x, y).

Let us recall Leibnitz’s theorem for the nth derivative of a product of functions; thus, we have

0

[ ] ( )( )n

n ax r ax n rr

r

D e nc D e Df f-

=

=Â

0

nax r n r

rr

e nc a D f-

=

Ê ˆ= Á ˜

Ë ¯Â

= eax(D + a)nfApplying this result, we arrive at

F(D, D¢)[eax+byf(x, y)] = eax+by F(D + a, D¢ + b) f(x, y) (1.83)

Hence, it follows that

1 1

[ ( , )] ( , )( , ) ( , )

ax by ax bye x y e x yF D D F D a D b

f f+ +=+ +¢ ¢

1( , )

( , )ax bye e x y

F D a Df= ◊

+ ¢

1( , )

( , )by axe e x y

F D D bf= ◊

+¢(1.84)



For illustration of various cases to find PI, here follows several examples:

EXAMPLE 1.19 Solve the equation (D2 + 3DD� + 2D�2)u = x + y.

Solution The given PDE is reducible, since it can be factored as

(D + D�)(D + 2D�)u = x + y (1)

Therefore,CF = �1(x – y) + �2(2x – y) (2)

where �1 and �2 are arbitrary functions.The PI of the given PDE is obtained as follows:

2 2

1PI ( )

( 3 2 )x y

D DD D� �

� ��

22

2

1( )

1 3 2

x yD D

DD D

� ��

12

2 2

11 3 2 ( )

D Dx y

DD D

� ��

2

11 3 ( )

Dx y

DD

��

2

1 3(1)x y

DD

� ��

2 3

2

1[ 2 ]

2 3

x xy x y

D� � � � (3)

Adding Eqs. (2) and (3), we have the complete solution of the given PDE as

2 3

1 2( ) ( 2 )2 3

x xu y x y x y� �� .

EXAMPLE 1.20 Solve the following equation (D – D� – 1)(D – D� – 2)u = e2x–y + x.

Solution The CF of the given PDE is

CF = ex�1(x + y) + e2x�2(x + y) (1)

The PI corresponding to the term e2x–y is

2 21 1

(2 1 1)(2 1 2) 2x y x ye e� ��

� � � �(2)



while the PI corresponding to the term x is

111

(1 ) 12 2 2

D DD D x

��

1(1 ) 1

2 2 2

D DD D x

� ��

1 1 1 3(1 )

2 2 2 2D D x x

� � � � � � � �� (3)

Combining Eqs. (1), (2) and (3), the complete solution of the given PDE is found to be

2 21 2

1 3( ) ( )

2 2 4x x x y x

u e x y e x y e� � �� .

EXAMPLE 1.21 Solve the following equation

(D2 + 2DD� + D�2 – 2D – 2D�)u = sin(x + 2y).

Solution The given PDE can be factored and rewritten as

(D + D�)(D + D� – 2)u = sin(x + 2y) (1)

for which the CF is given byCF = �1(x – y) + e2x�2(x – y) (2)

while

2 2

1PI sin( 2 )

( 2 2 2 )x y

D DD D D D� �

� � � �� .

Setting D2 = –1, DD� = –2, D�2 = –4, we get

1

PI sin( 2 )(2 2 9)

x yD D

� � ��

2 2

[2( ) 9]sin( 2 )

[4( 2 ) 81]

D Dx y

D DD D

� ��

2( ) 9sin( 2 )

117

D Dx y

� ��

1

117� [2 cos(x + 2y) + 4 cos(x + 2y) – 9 sin(x + 2y)]

1

39� [2 cos(x + 2y) – 3 sin(x + 2y)] (3)



Adding Eqs. (1), (2) and (3), we find that the complete solution of the given PDE as

u = �1(x – y) + e2x�2(x – y) + 1

39[2 cos(x + 2y) – 3 sin(x + 2y)].

EXAMPLE 1.22 Solve the following equation

(D2 – DD�)u = cos x cos 2y.


D(D – D�)u = cos x cos 2y (1)Its CF is given by

CF = �1(y) + �2(y + x), (2)while its PI is given by

2

1 1PI

2( )D DD� �

� �[cos(x + 2y) + cos(x – 2y)]

1 1 1

cos( 2 ) cos( 2 )2 ( 1 2) ( 1 2)

x y x y� ��

1 1

cos( 2 ) cos( 2 )2 6

x y x y� � � � (3)

Hence, the complete solution of the given PDE is given by

u = �1(y) + �2(y + x) + 1

2cos(x + 2y) –

1

6cos(x – 2y).

EXAMPLE 1.23 Find the solution of

(D2 + DD� – 6D�2)u = y cos x.


(D + 3D�)(D – 2D�)u = y cos x (1)

Its CF is given by

CF = �1(3x – y) + �2(2x + y) (2)

The PI of the given PDE is

1 1PI cos

( 3 ) ( 2 )y x

D D D D� �

� �� (3)

Applying the operator 1

( 2 )D D� � first on y cos x

(D – 2D�)u = y cos x



Its auxiliary equations are

1 2 cos

dx dy du

y x� ��

.

The first two members give

y + 2x = �� (constant).

From the first and the third members, we have

du = y cos x dx = (� – 2x) cos x dx,

on integration, we get

( 2 ) cosu x x dx�� where � is to be replaced by (y + 2x) after integration. Now, Eq. (3) gives

1

PI ( 2 ) sin sin ( 2)( 3 )

x x x dxD D

�� 1

[( 2 ) sin 2 cos ]( 3 )

x x xD D

��

1[ sin 2 cos ]

( 3 )y x x

D D� �

� �

[( 3 ) sin 2 cos ]x x x dx�� ( 3 )( cos ) 3 cos 2 sinx x x dx x�� .

Now, replacing � by (y – 3x), we get

PI = –y cos x + sin x (4)

Hence, the solution of the given PDE is found to be

u = CI + PI

= �1(3x – y) + �2(2x + y) – y cos x + sin x.

EXAMPLE 1.24 Show that a linear PDE of the type

( , )i j

i jij i j

i j

ua x y f x y

x y

��

can be reduced to a one with constant coefficients by the substitution

� � � = log x, = log y.



Solution

1u u u

x x x

xx x

∂ ∂ ∂ ∂= =∂ ∂ ∂ ∂

or u ux

x x∂ ∂=∂ ∂

That is,

(say)x Dx x∂ ∂= =∂ ∂

(1)

Therefore,

1 1

1 11 1

( 1)n n n

n n nn n n

u u ux x x n x

x x x x

- -- -

- -

Ê ˆ∂ ∂ ∂ ∂= + -Á ˜∂ ∂ ∂ ∂Ë ¯

or 1

11

1n n

n nn n

u ux x n x

xx x

--

-∂ ∂ ∂Ê ˆ= - +Á ˜Ë ¯∂∂ ∂

(2)

By setting n = 2, 3, 4, … in Eq. (2), we obtain

2

22

( 1) ( 1)u u

x D x D D uxx

∂ ∂= - = -

∂∂,

3

33

( 1)( 2)u

x D D D ux

∂= - -

∂,

and so on. Similarly, we can show that

u uy D u

y h∂ ∂= = ¢∂ ∂

,

2

22

( 1)u

y D D uy

∂ = -¢ ¢∂

,

and 2u

xy DD ux y

∂ = ¢∂ ∂

and so on. Substituting these results into the given PDE, it becomes

F(D, D¢)u = f(ex, eh) = f(x, h) (3)

where,

,D Dx h∂ ∂= =¢∂ ∂

Thus, Eq. (3) can be seen as a PDE with constant coefficients.



For illustration, here follows an example.

EXAMPLE 1.25 Solve the following PDE

(x2D2 + 2xy DD¢ + y2D¢2)u = x2y2 (1)

Solution Using the substitution

x = log x, h = log y and using the notation

,D Dx h∂ ∂= = ¢∂ ∂

respectively, the given PDE reduces to a PDE with constant coefficients, in the form

[D(D – 1) + 2DD¢ + D¢(D¢ – 1)]u = e2x+2h.

On rewriting, we have(D + D¢)(D + D¢ – 1)u = e2x+2h (2)

The CF of Eq. (2) is given byCF = f1(x – h) + ex f2 (x – h), (3)

while, its PI is obtained as

2 21PI

(2 2)(2 2 1)e x h+=

+ + -

2 21

12e x h+= . (4)

Transforming back from (x, h) to (x, y), we find the complete solution of the given PDE as

u = f1(log x – log y) + x f2(log x – log y) + 2 2

12

x y

or 2 21 2

1

12

x xu x x y

y yy yÊ ˆ Ê ˆ

= + +Á ˜ Á ˜Ë ¯ Ë ¯.

1.7 HOMOGENEOUS LINEAR PDE WITH CONSTANT COEFFICIENTS

Equation (1.61) is said to be linear PDE of nth order with constant coefficients. It is alsocalled homogeneous, because all the terms containing derivatives are of the same order. Now,Eq. (1.61) can be rewritten in operator notation as

[a0Dn + a1Dn–1 D¢ + a2Dn–2D¢2 + … + anD¢n]u = F(D, D¢)u = f(x, y) (1.85)

As in the case of ODE, the complete solution of Eq. (1.85) consists of the sum of CFand PI.



1.7.1 Finding the Complementary Function

Let us assume that the solution of the equation F(D, D¢)u = 0 in the form

u = f(y + mx) (1.86)

Then, Diu = mifi(y + mx), D¢

ju = f j(y + mx),

and DiD¢ju = mifi+j (y + mx).

Substituting these results into F(D, D¢)u = 0, we obtain

(a0mn + a1mn–1 + a2mn–2 + … + an)fn(y + mx) = 0,

which will be true, iff

a0mn + a1mn–1 + a2mn–2 + … + an = 0 (1.87)

This equation is called an auxiliary equation for F(D, D¢)u = 0.Let m1, m2, …, mn are the roots of Eq. (1.87). Depending on the nature of these roots,

several cases arise:

Case I When the roots m1, m2, …, mn are distinct. Corresponding to m = m1, the CF isu = f1(y + m1x). Similarly, u = f2(y + m2x), u = f3(y + m3x), etc. are all complementaryfunctions. Since, the given PDE is linear, using the principle of superposition, the CF ofEq. (1.85) can be written as

CF = f1(y + m1x) + f2(y + m2x) + … + fn(y + mnx)

where, f1, f2, …, fn are arbitrary.

Case II When some of the roots are repeated. Let two roots say m1 and m2 are repeated,and each equal to m. Consider the equation

(D – mD¢)(D – mD¢)u = 0.

Setting (D – mD¢)u = z, the above equation becomes

(D – mD¢)z = 0,

or 0z z

mx y

∂ ∂- =∂ ∂

which is of Lagrange’s form. Writing down its auxiliary equations, we have

1 0

dx dy dz

m= =-

.

The first two members gives y + mx = constant = a(say).The third member yields z = constant.Therefore, z = f(y + mx) is a solution.Substituting for z, we get

(D – mD¢)u = f(y + mx)



which is again in the Lagranges form, whose auxiliary equations are

1 ( )

dx dy du

m y mx��

which gives,

y + mx = constant

and u = x�(y + mx) + constant

Hence, the CF is u = x�(y + mx) + �(y + mx).

In general, if the root m is repeated r times, then the CF is given by

u = xr–1�1(y + mx) + xr–2�2(y + mx) + … + �r(y + mx)

where, �1, �2, …, �r are arbitrary.For illustration, we consider the following couple of examples.

EXAMPLE 1.26 Solve the following PDE (D3 – 3D2D� + 4D�3)u = 0.

Solution Observe that the given PDE is a linear homogeneous PDE. Dividing throughoutby D� and denoting (D/D�) by m, its auxiliary equation can be written as

m3 – 3m2 + 4 = (m + 1)(m – 2)2 = 0.

Therefore, the roots of the auxiliary equation are –1, 2, 2. Thus,

CF = �1(y – x) + �2(y + 2x) + x �3(y + 2x).


(D3 + DD�2 – 10D�3)u = 0.

Solution Observe that the given equation is a linear homogeneous PDE. Denoting (D/D�) by m. The auxiliary equation for the given PDE is given by

m3 + m – 10 = (m – 2)(m2 + 2m + 5) = 0.

Its roots are: 2, (–1 + 2i), (–1 – 2i).Hence, the required CF is found to be

u = �1(y + 2x) + �2(y – x + 2ix) + �3(y – x – 2ix).

��6�� "&/�03� �� 0��1� &/"��

Methods for finding the PI, in the case of linear homogeneous PDE’s are, similar to the one’sdeveloped in the Section 1.6 for linear non-homogeneous PDEs. That is, the PI for theequation

F(D, D�)u = f(x, y)is obtained from

uPI = PI = 1

( , )F D D� f(x, y).



However, when the above stated methods fail we have a general method, which is applicablewhatever may be the form of f(x, y), and is presented below:

We have already assumed that F(D, D¢) can be factorised, in general, say into n linearfactors. Thus,

1PI ( , )

( , )f x y

F D D=

¢

1 2

1PI ( , )

( )( ) ( )n

f x yD m D D m D D m D

=- - -¢ ¢ ¢K

1 2

1 1 1( , )

( ) ( ) ( )n

f x yD m D D m D D m D

= ◊- - -¢ ¢ ¢

L .

In general, to evaluate

1( , )

( )f x y

D mD- ¢,

we consider the equation

(D – mD¢)u = f(x, y)

or p – mq = f(x, y) (Lagrange’s form)

for which the auxiliary equations are

1 ( , )

dx dy du

m f x y= =-

.

Its first two members, yield

y + mx = c (constant)

The first and last members gives us

du = f(x, y)dx = f(x, c – mx).


= -Ú ( , )u f x c mx dx

or 1( , ) ( , )

( )f x y f x c mx dx

D mD= -

- ¢ ÚAfter integration, we shall immediately replace c by (y + mx). Applying this procedurerepeatedly, we can find the PI for the given PDE. For illustration, we consider the followingexamples:


(D2 – 4DD¢ + 4D¢2)u = e2x+y.



Solution The given equation is a linear homogeneous PDE. Its auxiliary equation canbe written as

m2 – 4m + 4 = (m – 2)2 = 0.

In this example, the roots are repeated and they are 2, 2. The complementary function andparticular integral are obtained as

CF = f1(y + 2x) + xf2(y + 2x) (1)

and 22

1PI

( 2 )x ye

D D+=

- ¢

If we set D = 2, D¢ = 1, we observe that F(D, D¢) = 0, which is a failure case. Therefore,we shall adopt the general method to find PI. Now, noting that

1( , ) ( , )

( )f x y f x c mx dx

D mD= -

- ¢ Ú .

21 1PI

( 2 ) ( 2 )x ye

D D D D+= ◊

- -¢ ¢

(2 2 )1

( 2 )x c xe dx

D D+ -=

- ¢ Ú21 1

( 2 ) ( 2 )c y xxe xe

D D D D+= =

- -¢ ¢

( 2 2 )c x x cxe dx e xdx- += =Ú Ú2 2

2

2 2c y xx x

e e += = (2)

From Eqs. (1) and (2), the complete solution of the given PDE is found to be

u = f1(y + 2x) + xf2(y + 2x) + 2

2

xey+2x.

EXAMPLE 1.29 Find a real function u(x, y), which reduces to zero when y = 0 and satisfythe PDE

2 22 2

2 2( )

u ux y

x yp∂ ∂+ = - +

∂ ∂.

Solution In symbolic form, the given PDE can be written as

(D2 + D¢2)u = –p(x2 + y2)

Its auxiliary equation is given by

(m2 + 1) = 0, which gives m = ±i.



Hence,CF = f1(y + ix) + f2(y – ix) (1)

andpp- += + = -

+ +¢ ¢ ¢

2 22 2

2 2 2 2 2

1 ( )PI ( )

( ) (1 / )

x yx y

D D D D D

122 2

2 21 ( )

Dx y

D D

p-Ê ˆ

= - + +Á ˜¢ ¢Ë ¯

22 2

2 21 ( )

Dx y

D D

p È ˘= - - + +Í ˙

¢ ¢Í ˙Î ˚L

2 22 4

2( )x y

D D

p p= - + +¢ ¢

32

3

2

3

yx y y

D D

p pÊ ˆ= - + +Á ˜¢ ¢Ë ¯

or 2 2 4 2

2

2PI

2 12 2

x y y y

D

ppÊ ˆ

= - + +Á ˜ ¢Ë ¯

2 2 4 4

22 12 24

x y y yp pÊ ˆ

= - + +Á ˜Ë ¯

2 2

2x y

p= - (2)

Hence, the complete solution of the given PDE is found to be

u = f1(y + ix) + f2(y – ix) – 2

px2y2 (3)

Finally, the real function satisfying the given PDE is given by

2 2

2u x y

p= - (4)

which of course Æ 0 as y Æ 0.

EXERCISES

1. Find the region in the xy-plane in which the following equation is hyperbolic:2 2[( ) 1] 2 [( ) 1] 0xx xy yyx y u u x y u− − + + − − =



2. Find the families of characteristics of the PDE2(1 ) 0xx yyx u u� � �

in the elliptic and hyperbolic cases.3. Reduce the following PDE to a canonical form

0xx yyu xyu� �

4. Classify and reduce the following equations to a canonical form:

(a) 2 2 0,xx yyy u x u� � 0,x � 0.y �

(b) 2 0.xx xy yyu u u� � �

(c) .x yxx yye u e u u� �

(d) 2 22 0.xx xy yyx u xyu y u� � �

(e) 4 5 2.xx xy yy x yu u u u u� � � � �5. Reduce the following equation to a canonical form and hence solve it:

3 10 3 0xx xy yyu u u� � �

6. If 2( ) ,xx ttL u c u u� � then show that its adjoint operator is given by

2* xx ttL c v v� �

7. Determine the adjoint operator L* corresponding to

( ) xx xy yy x yL u Au Bu Cu Du Eu Fu� � � � � �

where A, B, C, D, E and F are functions of x and y only.8. Find the solution of the following Cauchy problem

( , )xyu F x y�

given

( ),u f x� ( )u

g xn

��

� on the line y x�

using Riemann’s method which is of the form

0

00 0 0 0

IR

1 1( , ) [ ( ) ( )] ( ) ( , )

2 2

y

xu x y f x f y g x dx F x y dx dy� � � ��

where IR is the triangular region in the xy-plane bounded by the line y x� and the

lines 0 0,x x y y� � through 0 0( , ).x y



9. The characteristics of the partial differential equation

2 2 22

2 22 cos 2 3 0

z z z z zx

x y x yx y

� � � � ��

� � � � �

when it is of hyperbolic type are… and… (GATE-Maths, 1997)

10. Using x yη � � as one of the transformation variable, obtain the canonical form of

2 2 2

2 22 0.

u u u

x yx y

� � ��

� � �

(GATE-Maths, 1998)

Choose the correct answer in the following questions (11–14):11. The PDE

3 2( 1) 0xx yyy u x u� � � is

(A) parabolic in {( , ) : 0}x y x �

(B) hyperbolic in {( , ) : 0}x y y �

(C) elliptic in 2IR

(D) parabolic in {( , ) : 0}.x y x � (GATE-Maths, 1998)

12. The equation2 2 2( 1) ( 1) ( 1) 0xx xy yy xx y z x y z y y z z� � � � � � �

is hyperbolic in the entire xy-plane except along(A) x-axis (B) y-axis(C) A line parallel to y-axis (D) A line parallel to x-axis.

(GATE-Maths, 2000)13. The characteristic curves of the equation

2 2 2 2 ,xx yyx u y u x y x� � � 0x � are

(A) rectangular hyperbola (B) parabola(C) circle (D) straight line.

(GATE-Maths, 2000)14. Pick the region in which the following PDE is hyperbolic:

2xx xy yy x yyu xyu xu u u� � � �

(A) 1xy � (B) 0xy �

(C) 1xy � (D) 0.xy �(GATE-Maths, 2003)



15. Solve the following PDEs:

(i) (D – D� – 1)(D – D� – 2)u = 0

(ii) (D + D� – 1)(D + 2D� – 3)u = 0

16. Solve the following PDE:(D2 + DD� + D + D� + 1)u = 0

17. Solve the following PDEs:

(i) (D2 – DD� + D� – 1)u = cos(x + 2y)

(ii) D(D – 2D� – 3)u = ex+2y

(iii) (2D + D� – 1)2(D – 2D� + 2)3 = 0

18. Find the complete solution of the following PDEs:

(i) (x2D2 – 2xy DD� + y2D�2 – xD + 3yD�)u = 8(y/x)

(ii) D(D – 2D� – 3)u = ex+2y

19. Find the complete solution of the following PDEs:

(i) (D2 + 3DD� + 2D�2)u = x + y

(ii) (D2 + D�2)u = cos px cos qy

(iii) (D2– DD� – 2D�2)u = (y – 1)ex

(iv) (4D2 – 4DD� + D�2)u = 16 log (x + 2y)

(v) (D2 – 3DD� + 2D�2)u = e2x+3y + sin (x – 2y)


��

In Chapter 1, we have seen the classification of second order partial differential equation intoelliptic, parabolic and hyperbolic types. In this chapter we shall consider various propertiesand techniques for solving Laplace and Poisson equations which are elliptic in nature.

Various physical phenomena are governed by the well known Laplace and Poisson equations.A few of them, frequently encountered in applications are: steady heat conduction, seepagethrough porous media, irrotational flow of an ideal fluid, distribution of electrical and magneticpotential, torsion of prismatic shaft, bending of prismatic beams, distribution of gravitationalpotential, etc. In the following two sub-sections, we shall give the derivation of Laplace andPoisson equations in relation to the most frequently occurring physical situation, namely, thegravitational potential.

�� !�� "#��

Consider two particles of masses m and m1 situated at Q and P separated by a distance r asshown in Fig. 2.1. According to Newton’s universal law of gravitation, the magnitude of theforce, proportional to the product of their masses and inversely proportional to the square ofthe distance, between them is given by

12

�mm

F Gr

(2.1)

where G is the gravitational constant. It r represents the vector ,��

PQ assuming unit mass at

Q and 1,�G the force at Q due to the mass at P is given by

1 13

� �� m m

rr

rF (2.2)

106

��

Elliptic Differential Equations


ELLIPTIC DIFFERENTIAL EQUATIONS 107

z

y

x

OS

Pm1

�

r

Q m( )

Fig. 2.1 Illustration of Newton’s universal law of gravitation.

which is called the intensity of the gravitational force. Suppose a particle of unit mass movesunder the attraction of a particle of mass m1 at P from infinity up to Q; then the work doneby the force F is

1 1

� �

� �� r r m m

d dr r

F r r (2.3)

This is defined as the potential V at Q due to a particle at P and is denoted by

1� �m

Vr

��

From Eq. (2.2), the intensity of the force at P is

� ��VF (2.5)

Now, if we consider a system of particles of masses 1 2, , ,� nm m m which are at distances

1 2, , ,� nr r r respectively, then the force of attraction per unit mass at Q due to the system is

1 1

n ni i

i ii i

m m

r r� �� F (2.6)

The work done by the force acting on the particle is

1� �

� � � ��nr

i

ii

md V

rF r (2.7)



Therefore,

2 2 2

1 1

0,� �

� � �� n n

i i

i ii i

m mV

r r0�ir (2.8)

where

2 2 22

2 2 2div

x y z� � � � � ��

� � �is called the Laplace operator.

In the case of continuous distribution of matter of density ρ in a volume ,τ we have

( , , )( , , )

τ

ρ ξ η ζ τ� ��V x y z dr

(2.9)

where 2 2 2 1/2{( ) ( ) ( ) }ξ η ζ� � � � � �r x y z and Q is outside the body. It can be verified that

2 0� �V (2.10)

which is called the Laplace equation.

�� $$�� "#��

Consider a closed surface S consisting of particles of masses 1 2, , , .� nm m m Let Q be any

point on S. Let1

n

ii

m M�

�Σ be the total mass inside S, and let 1 2, , ,� ng g g be the gravity

field at Q due to the presence of 1 2, , ,� nm m m respectively within S. Also, let1

,n

ii

g g�

�Σ the

entire gravity field at Q. Then, according to Gauss law, we have

4π� � ��S

d GMg S (2.11)

where ,M d� ��τ

ρ τ ρ is the mass density function and τ is the volume in which the masses

are distributed throughout. Since the gravity field is conservative, we have

� �Vg (2.12)

where V is a scalar potential. But the Gauss divergence theorem states that

τ

τ� � � �� S

d dg S g (2.13)

Also, Eq. (2.11) gives

4τ

π ρ τ� � �� S

d G dg S (2.14)



Combining Eqs. (2.13) and (2.14), we have

( 4 ) 0τ

π ρ τ∇ ⋅ + =∫∫∫ G dg

implying

4π ρ∇ ⋅ = − = ∇ ⋅∇G Vg

Therefore,

2 4π ρ∇ = −V G (2.15)

This equation is known as Poisson’s equation.

2.2 BOUNDARY VALUE PROBLEMS (BVPs)

The function V, whose analytical form we seek for the problems stated in Section 2.1, in

addition to satisfying the Laplace and Poisson equations in a bounded region IR in R3, should

also satisfy certain boundary conditions on the boundary IR∂ of this region. Such problems

are referred to as boundary value problems (BVPs) for Laplace and Poisson equations. We

shall denote the set of all boundary points of IR by IR .∂ By the closure of IR, we mean the

set of all interior points of IR together with its boundary points and is denoted

by IR. Symbolically, IR. IR IR .U∂=

If a function ( )∈ nf c ( f “belongs to” c(n)), then all its derivatives of order n are continuous.

If it belongs to c(0), then we mean f is continuous.

There are mainly three types of boundary value problems for Laplace equation. If (0)∈f c and

is specified on the boundary IR∂ of some finite region IR, the problem of determining a

function ( , , )ψ x y z such that 2 0ψ∇ = within IR and satisfying ψ = f on IR∂ is called the

boundary value problem of first kind, or the Dirichlet problem. For example, finding the

steady state temperature within the region IR when no heat sources or sinks are present and

when the temperature is prescribed on the boundary IR,∂ is a Dirichlet problem. Another

example would be to find the potential inside the region IR when the potential is specified

on the boundary IR .∂ These two examples correspond to the interior Dirichlet problem.

Similarly, if (0)∈f c and is prescribed on the boundary IR∂ of a finite simply

connected region IR, determining a function ( , , )ψ x y z which satisfies 2 0ψ∇ = outside IR and

is such that ψ = f on IR,∂ is called an exterior Dirichlet problem. For example, determination

of the distribution of the potential outside a body whose surface potential is prescribed, is an



exterior Dirichlet problem. The second type of BVP is associated with von Neumann. The

problem is to determine the function ( , , )ψ x y z so that 2 0ψ� � within IR while / nψ� � is

specified at every point of IR,� where / n�� denotes the normal derivative of the field

variable .ψ This problem is called the Neumann problem. If ψ is the temperature, / nψ� � is

the heat flux representing the amount of heat crossing per unit volume per unit time alongthe normal direction, which is zero when insulated. The third type of BVP is concerned with

the determination of the function ( , , )ψ x y z such that 2 0ψ� � within IR, while a boundary

condition of the form / ,n h f� � � �ψ ψ where 0�h is specified at every point of IR .� This

is called a mixed BVP or Churchill’s problem. If we assume Newton’s law of cooling, the

heat lost is ,ψh where ψ is the temperature difference from the surrounding medium and 0�h is

a constant depending on the medium. The heat f supplied at a point of the boundary is partlyconducted into the medium and partly lost by radiation to the surroundings. Equating theseamounts, we get the third boundary condition.

��+ ��(� �(�� (� �(� ��

Among the mathematical tools we employ in deriving many important results, the Gauss

divergence theorem plays a vital role, which can be stated as follows: Let IR� be a closed

surface in the xyz-space and IR denote the bounded region enclosed by IR� in which F is

a vector belonging to c(1) in IR and continuous on IR. Then

IR IR

n̂ dS dV�

� � � �� F F (2.16)

where dV is an element of volume, dS is an element of surface area, and n̂ the outward drawnnormal.Green’s identities which follow from divergence theorem are also useful and they can be

derived as follows: Let ,φ�F f where f is a vector function of position and φ is a scalar

function of position. Then,

IR IR

ˆ( ) dV n dS�

� � � �� f fφ φ

Using the vector identity

( )φ φ φ� � � �� f f f

we have

IR IR IR

ˆdV n dS dV�

�� f f fφ φ φ



If we choose ,ψ� �f the above equation yields

2

IR IR IR

ˆdV n dS dV�

� �� φ ψ φ ψ φ ψ ��

Noting that ˆ ψ��n is the derivative of ψ in the direction of ˆ,n we introduce the notation

ˆ /n n� �� ψ ψ

into Eq. (2.17) to get

2

IR IR IR

dV dS dVn

�

��

� �� ψφ ψ φ φ ψ ��

This equation is known as Green’s first identity. Of course, it is assumed that

both φ and ψ possess continuous second derivatives.

Interchanging the role of φ and ,ψ we obtain from relation (2.18a) the equation

2

IR IR IR

dV dS dVn

�

��

� �� φψ φ ψ ψ φ ��

Now, subtracting Eq. (2.18b) from Eq. (2.18a), we get

2 2

IR IR

( )dV dSn n

�

� ��

� � � � � � �� ψ φφ ψ ψ φ φ ψ ��

This is known as Green’s second identity. If we set φ ψ� in Eq. (2.18a) we get

2 2

IR IR IR

( ) dV dS dVn

�

��

� � � �� φφ φ φ φ ��

which is a special case of Green’s first identity.

��, �� (��

Solutions of Laplace equation are called harmonic functions which possess a number ofinteresting properties, and they are presented in the following theorems.

Theorem 2.1 If a harmonic function vanishes everywhere on the boundary, then it is identicallyzero everywhere.

Proof If φ is a harmonic function, then 2 0φ� � in IR . Also, if 0φ � on IR,� we

shall show that 0φ � in IR IR IR .U� � Recalling Green’s first identity, i.e., Eq. (2.20), we get

2 2

IR IR IR

( ) dV dS dVn

�

��

� � � �� φφ φ φ φ



and using the above facts we have, at once, the relation

2

IR

( ) 0dV� �� φ

Since 2( )φ� is positive, it follows that the integral will be satisfied only if 0.φ� � This

implies that φ is a constant in IR. Since φ is continuous in IR and φ is zero on IR,� it

follows that 0φ � in IR .

Theorem 2.2 If φ is a harmonic function in IR and / 0nφ �� on IR,� then φ is a constant

in IR.

Proof Using Green’s first identity and the data of the theorem, we arrive at

2

IR

( ) 0dV� �� φ

implying 0,φ� � i.e., φ is a constant in IR . Since the value of φ is not known on the boundary

IR� while / 0,n� � �φ it is implied that φ is a constant on IR� and hence on IR.

Theorem 2.3 If the Dirichlet problem for a bounded region has a solution, then it is unique.

Proof If 1φ and 2φ are two solutions of the interior Dirichlet problem, then

21 1

22 2

0 in IR; on IR

0 in IR; on IR

f

f

�

�

� � �

� � �

φ φ

φ φ

Let 1 2.ψ φ φ� � Then

2 2 21 2

1 2

0 in IR;

0 on IRf f �

� � � ��

� � � � �

ψ φ φ

ψ φ φTherefore,

2 0 in IR,� �ψ 0ψ � on IR�

Now using Theorem 2.1, we obtain 0ψ � on IR, which implies that 1 2.φ φ� Hence, the solutionof the Dirichlet problem is unique.

Theorem 2.4 If the Neumann problem for a bounded region has a solution, then it is eitherunique or it differs from one another by a constant only.



Proof Let 1φ and 2φ be two distinct solutions of the Neumann problem. Then we have

2 11

2 22

0 in IR; on IR,

0 in IR; on IR

fn

fn

∂ ∂∂∂ ∂∂

∇ = =

∇ = =

φφ

φφ

Let 1 2.ψ φ φ= − Then

2 2 21 2

1 2

0 in IR

0 on IRn n n

∂ ∂∂ ∂∂ ∂ ∂

∇ = ∇ − ∇ =

= − =

ψ φ φ

φ φψ

Hence from Theorem 2.2, ψ is a constant on IR, i.e., 1 2φ φ− = constant. Therefore, the solutionof the Neumann problem is not unique. Thus, the solutions of a certain Neumann problemcan differ from one another by a constant only.

2.4.1 The Spherical Mean

Let IR be a region bounded by IR∂ and let ( , , )P x y z be any point in IR . Also,

let ( , )S P r represent a sphere with centre at P and radius r such that it lies entirely within

the domain IR as depicted in Fig. 2.2. Let u be a continuous function in IR . Then thespherical mean of u denoted by u is defined as

2( , )

1( ) ( )4 S P r

u r u Q dSr

= ∫∫π(2.21)

x

y

z

Pr

S P r( , )

Q ( , , )ζξ η

IR

O

∂IR

Fig. 2.2 Spherical mean.



where ( , , )Q ξ η ζ is any variable point on the surface of the sphere ( , )S P r and dS is the

surface element of integration. For a fixed radius r, the value ( )u r is the average of the values

of u taken over the sphere ( , ),S P r and hence it is called the spherical mean. Taking the

origin at P, in terms of spherical polar coordinates, we have

sin cos

sin sin

cos

x r

y r

z r

ξ θ φ

η θ φ

ζ θ

� �

� �

� �

Then, the spherical mean can be written as

2 22 0 0

1( ) ( sin cos , sin sin , cos ) sin

4u r u x r y r z r r d d

r

π π

φ θθ φ θ φ θ θ θ φ

π � ��

Also, since u is continuous on ( , ),S P r u too is a continuous function of r on some interval

0 ,r R� � which can be verified as follows:

2

0 0

1 ( )( ) ( ) sin sin ( )

4 4

u Qu r u Q d d d d u Q

π πθ θ φ θ θ φ

π π� � ��

Now, taking the limit as 0, ,r Q P� � we have

0Lt ( ) ( )

ru r u P

�� (2.22)

Hence, u is continuous in 0 .r R� �

��,�� (�� '� #�� -��.� �� .��!� �#�!��$

Theorem 2.5 Let u be harmonic in a region IR . Also, let P(x, y, z) be a given point in IRand S(P, r) be a sphere with centre at P such that S(P, r) is completely contained in the domainof harmonicity of u. Then

2( , )

1( ) ( ) ( )

4S P r

u P u r u Q dSr

� � ��π

Proof Since u is harmonic in IR, its spherical mean ( )u r is continuous in IR and isgiven by

2 22 2 0 0

( , )

1 1( ) ( ) ( , , ) sin

4 4S P r

u r u Q dS u r d dr r

� �� π π

ξ η ζ θ θ φπ π



Therefore,

2

0 0

2

0 0

( ) 1( ) sin

4

1( sin cos sin sin cos ) sin

4

r r rdu r

u u u d ddr

u u u d d

π πξ η ζ

π πξ η ζ

ξ η ζ θ θ φπ

θ φ θ φ θ θ θ φπ

� � �

� � �

� �

� � (2.23)

Noting that sin cos , sin sinθ φ θ φ and cos θ are the direction cosines of the normal n̂ on

S(P, r),

,u iu ju kuξ η ζ� � � � 1 2 3ˆ ( , , ),n in jn kn�

the expression within the parentheses of the integrand of Eq. (2.23) can be written as ˆ.u n� � Thus

22

( , )

2( , )

2( , )

22

( , )

( ) 1ˆ sin

4

1ˆ

4

1(by divergence theorem)

4

10 (since is harmonic)

4

S P r

S P r

V P r

V P r

du ru nr d d

dr r

u n dSr

u dVr

u dV ur

� � �

� � �

� � ��

� � �

��

��

��

��

θ θ φπ

π

π

π

Therefore, 0,du

dr� implying u is constant.

Now the continuity of u at 0r � gives, from Eq.(2.22), the relation

2( , )

1( ) ( ) ( )

4S P r

u r u P u Q dSr

� � ��π(2.24)

��,�+ (�/�.#.0(��.#.� ��!�� 1� ��$�"#��!�$

Theorem 2.6 Let IR be a region bounded by IR .� Also, let u be a function which is

continuous in a closed region IR and satisfies the Laplace equation 2 0u� � in the interiorof IR . Further, if u is not constant everywhere on IR, then the maximum and minimum valuesof u must occur only on the boundary IR� .



Proof Suppose u is a harmonic function but not constant everywhere on IR. If possible,

let u attain its maximum value M at some interior point P in IR . Since M is the maximum

of u which is not a constant, there should exist a sphere ( , )S P r about P such that some of

the values of u on ( , )S P r must be less than M. But by the mean value property, the value

of u at P is the average of the values of u on ( , ),S P r and hence it is less than M. This

contradicts the assumption that u M� at P. Thus u must be constant over the entire

sphere ( , ).S P r

Let Q be any other point inside IR which can be connected to P by an arc lying entirely

within the domain IR . By covering this arc with spheres and using the Heine-Borel theoremto choose a finite number of covering spheres and repeating the argument given above, wecan arrive at the conclusion that u will have the same constant value at Q as at P. Thus u

cannot attain a maximum value at any point inside the region IR. Therefore, u can attain its

maximum value only on the boundary IR .� A similar argument will lead to the conclusion

that u can attain its minimum value only on the boundary IR .�Some important consequences of the maximum-minimum principle are given in the following

theorems.

Theorem 2.7 (Stability theorem). The solutions of the Dirichlet problem depend continuouslyon the boundary data.

Proof Let u1 and u2 be two solutions of the Dirichlet problem and let f1 and f2 be given

continuous functions on the boundary IR� such that

21 1 1

22 2 2

0 in IR; on IR,

0 in IR; on IR

u u f

u u f

�

�

� � �

� � �

Let 1 2.u u u� � Then,

2 2 21 2 1 20 in IR; on IRu u u u f f ��

Hence, u is a solution of the Dirichlet problem with boundary condition 1 2u f f� � on IR .� By

the maximum-minimum principle, u attains the maximum and minimum values on IR .� Thus

at any interior point in IR, we shall have, for a given 0,ε �

min maxu u uε ε� � � � �

Therefore,

| |u ε� in IR, implying 1 2| |u u ε� �



Hence, if

1 2| |f f ε− < on IR,∂ then 1 2| |u u ε− < on IR

Thus, small changes in the initial data bring about an arbitrarily small change in thesolution. This completes the proof of the theorem.

Theorem 2.8 Let {fn} be a sequence of functions, each of which is continuous on IR and

harmonic on IR . If the sequence {fn} converges uniformly on IR,∂ then it converges uniformly

on IR.

Proof Since the sequence {fn} converges uniformly on IR,∂ for a given 0,ε > we

can always find an integer N such that

| | for ,n mf f n m N− < >ε

Hence, from stability theorem, for all , ,n m N> it follows immediately that

| | in IRn mf f− < ε

Therefore, {fn} converges uniformly on IR.

EXAMPLE 2.1 Show that if the two-dimensional Laplace equation 2 0u∇ = is transformed

by introducing plane polar coordinates ,r θ defined by the relations cos , sin ,x r y rθ θ= = it

takes the form

2 2

2 2 21 1 0u u ur rr r

∂ ∂ ∂∂∂ ∂

+ + =θ

Solution In many practical problems, it is necessary to write the Laplace equation

in various coordinate systems. For instance, if the boundary of the region IR∂ is a circle,

then it is natural to use polar coordinates defined by cos , sin .x r y rθ θ= = Therefore,

2 2 2 ,r x y= + 1tan ( / )y xθ −=

cos ,xr θ= sin ,yr θ= sin ,x rθθ = − cos

y rθθ =

since

( , )u u r θ=sincosx r x x ru u r u u u

rθ θθθ θ⎛ ⎞= + = −⎜ ⎟⎝ ⎠

Similarly,

cossiny r y y ru u r u u urθ θ

θθ θ⎛ ⎞= + = +⎜ ⎟⎝ ⎠



Now for the second order derivatives,

sin sin sin( ) ( ) ( ) cos cos cosxx x x x r x x x r r

r

u u u r u u u u ur r rθ θ θ

θ

θ θ θθ θ θ θ� � � � � � � � � � � � � ��

Therefore,

2

sin sincos cos

sin cos sincos sin

xx rr r

r r

u u u ur r

u u u ur r r

θ θ

θ θθ θ

θ θθ θ

θ θ θθ θ

� ��

� � � �� (2.25)


2

cos cossin sin

cos sin cossin cos

yy rr r

r r

u u u ur r

u u u ur r r

θ θ

θ θθ θ

θ θθ θ

θ θ θθ θ

� ��

� � � �� (2.26)

By adding Eqs. (2.25) and (2.26) and equating to zero, we get

2

1 10xx yy rr ru u u u u

r rθθ� � � � � (2.27)

which is the Laplace equation in polar coordinates. One can observe that the Laplace equationin Cartesian coordinates has constant coefficients only, whereas in polar coordinates, it hasvariable coefficients.

EXAMPLE 2.2 Show that in cylindrical coordinates , ,r zθ defined by the relations cos ,x r θ�

sin , ,y r z zθ� � the Laplace equation 2 0u� � takes the form

2 2 2

2 2 2 2

1 10

u u u u

r rr r z

� � � ��

� � � �θ

Solution The Laplace equation in Cartesian coordinates is

2 2 22

2 2 20

u u uu

x y z

� � ��

� � � � �

The relations between Cartesian and cylindrical coordinates give

2 2 2 ,r x y� � 1tan ( / ),y xθ �� z z�



Since

( , , )

sincos

cossin

� �

� �

�

�

��

��

�

�

� ��

� ��

� � � �

x r x x z x r

y r y y z y r

z r z z z z

u u r z

u u r u u z u ur

u u r u u z u ur

u u r u u u

for the second order derivatives, we find

2

( ) ( ) ( ) ( )

sin sin sincos cos cos

sin sincos cos

sin cos sincos sin

xx x x x r x x x x z x

r rr

rr r

r r

u u u r u u z

u u u ur r r

u u ur r

u u u ur r r

� � � �

� � � ��

� ��

� � � ��

θ

θ θθ

θ θ

θ θθ θ

θ

θ θ θθ θ θ

θ θθ θ

θ θ θθ θ ��

Similarly

2

( ) ( ) ( ) ( )

cos cos cossin sin sin

cos cossin sin

cos sin cossin cos

yy y y y r y y y y z y

r rr

rr r

r r

zz zz

u u u r u u z

u u u ur r r

u u ur r

u u u ur r r

u u

θ

θ θθ

θ θ

θ θθ θ

θ

θ θ θθ θ θ

θ θθ θ

θ θ θθ θ

� � � �

� � � � � � � � � ��

� � � � ��

� � � � � � � ��

�

(2.29)

(2.30)

Adding Eqs. (2.28)–(2.30), we obtain

22

1 1rr r zzu u u u u

r rθθ� � � � � (2.31)



EXAMPLE 2.3 Show that in spherical polar coordinates , ,r θ φ defined by the relations

sinx r� θ cos , sin sin , cos ,y r z rφ θ φ θ� � the Laplace equations 2 0u� � takes the form

22

2 2

1 1sin 0

sin sin

u u ur

r r

� � � � ��

� � � �� θ

θ θ θ θ φ

Solution In Cartesian coordinates, the Laplace equation is

2 0xx yy zzu u u u� � � � �

In spherical coordinates, 2 2 2 2( , , ), , cos / , tan / .u u r r x y z z r y xθ φ θ φ� � � � � �It can be easily verified that

cos cos,x r

θ φθ � cos sin,y r

θ φθ � sinz r

θθ � �

sin,

sinx r

φφθ

� � cos,

siny r

φφθ

� 0zφ �

Now,

cos cos sinsin cos

sin

coscos sinsin sin

sin

sincos

x r x x x r

y r y y y r

z r z z z r

u u r u u u u ur r

uu u r u u u u

r r

u u r u u u ur

θ φ θ φ

φθ φ θ

θ φ θ

θ φ φθ φ θ φθ

φθ φθ φ θ φθ

θθ φ θ

� � � � � �

� � � � � �

� ��

For the second order derivatives,

( ) ( ) ( )

cos cos sinsin cos (sin cos )

sin

cos cos sin cos cossin cos

sin

cos cos sin sinsin cos

sin sin

xx x r x x x x x

rr

r

r

u u r u u

u u ur r

u u ur r r

u u ur r r

θ φ

θ φ

θ φθ

θ φφ

θ φ

θ φ φθ φ θ φθ

θ φ φ θ φθ φθ

θ φ φ φθ φθ θ

� � �

� ��

� � � ��

� � � ��



2 2 22 2

2 2 2

2

2 2 2

2

2

2 2 2 2 2

cos cos sin(sin cos )

sin

2 sin cos cos 2 sin cos

2 cos cos sin cos cos sin

sin

sin cos cos cos sin sin cos

sin sin

rr

r r

r

u u ur r

u ur r

u ur rr

ur r r

θθ φφ

θ φ

θφ

φ

θ φ φθ φθ

θ θ φ φ φ

θ φ φ θ φ φθ

φ φ θ φ φ φ φθ

� � �

� � � � � � � � �

� � � � � � � ��

� � �

2 2

2 2

cos sin 2 cos sin cos

sinu

r rθ

θ

θ φ θ θ φθ

� ��

� � � ��

��

2 2

cos sin cos( ) ( ) ( ) sin sin (sin sin )

sin

cos sin cos cos sinsin sin

sin

cos sin cos cossin sin

sin sin

co(sin sin )

yy y r y y y y y rr

r

r

rr

u u r u u u u ur r

u u ur r r

u u ur r r

u

θ φ θ φ

θ φθ

θ φφ

θ φ φθ φ θ φ θ φθ

θ φ φ θ φθ φθ

θ φ φ φθ φθ θ

θ φ

� � � � � � � ��

� � � � ��

� � � � ��

� �2 2 2

2 2 2

2

2 2 2

2

2 2

2 2

2 2

s sin cos

sin

2 sin cos sin 2 cos sin

2 cos cos sin cos sin cos

sin

2 sin cos sin cos cos

sin

sin cos sin cos

sin

r r

r

u ur r

u ur r

u ur rr

ur r

ur r

θθ φφ

θ φ

θφ

θ

φ

θ φ φθ

θ θ φ φ φ

θ φ φ θ φ φθ

θ θ φ θ φθ

φ φ φ φ

�

� � � � � � � �

� � � � � � ��

� � � � ��

� � �2

2 2 2

cos sin cos

sinr

θ φ φθ θ

� � ��

��



Similarly,

22

2

2

2

( ) ( ) ( )

sin sin sincos (cos ) cos

2 sin cos sincos

sin cos sin

zz z r z z z z z

r rr

rr r

r

u u r u u

u u u ur r r

u u ur r

u ur r

θ φ

θ θθ

θ θθ

θ

θ φ

θ θ θθ θ θ

θ θ θθ

θ θ θ

� � �

� � � � � � � � � � � ��

� � �

� � ��

Adding Eqs. (2.32)–(2.34), we obtain

22 2 2 2

1 1 2 cos0

sin sinrr ru u u u u u

rr r rθθ φφ θ

θθ θ

� � � � � � �


22 2

2 2

1 1sin 0

sin sin

u u uu r

r r

� � � � ��

� � � �� θ

θ θ θ θ φ(2.35)

��2 �� '��%��

The method of separation of variables is applicable to a large number of classical linearhomogeneous equations. The choice of the coordinate system in general depends on the shapeof the boundary. For example, consider a two-dimensional Laplace equation in Cartesiancoordinates.

2 0xx yyu u u� � � � (2.36)

We assume the solution in the form

( , ) ( ) ( )u x y X x Y y� (2.37)

Substituting in Eq. (2.36), we get

0X Y Y X� �� i.e.

X Yk

X Y

��

where k is a separation constant. Three cases arise.



Case I Let 2 ,k p p� is real. Then

22

20

d Xp X

dx� � ��

22

20

d Yp Y

dy� �

whose solution is given by

1 2px pxX c e c e��

and

3 4cos sinY c py c py�

Thus, the solution is

1 2 3 4( , ) ( ) ( cos sin )px pxu x y c e c e c py c py�� (2.38)

Case II Let 0.k � Then

2

20

d X

dx� ��

2

20

d Y

dy�

Integrating twice, we get

5 6X c x c� �

and

7 8Y c y c�

The solution is therefore,

5 6 7 8( , ) ( ) ( )u x y c x c c y c� (2.39)

Case III Let 2.k p� � Proceeding as in Case I, we obtain

9 10

11 12

cos sin

py py

X c px c px

Y c e c e�

� �

� �

Hence, the solution in this case is

9 10 11 12( , ) ( cos sin ) ( )py pyu x y c px c px c e c e�� (2.40)

In all these cases, ( 1, 2, ,12)ic i � � refer to integration constants, which are calculated by

using the boundary conditions.



��3 �� %�(�� 4�

The Dirichlet problem for a rectangle is defined as follows:

2PDE: 0, 0 , 0

BCs: ( , ) ( , ) 0, (0, ) 0, ( , 0) ( )

u x a y b

u x b u a y u y u x f x

� � � � � �

� � � � (2.41)

This is an interior Dirichlet problem. The general solution of the governing PDE, using themethod of variables separable, is discussed in Section 2.5. The various possible solutions ofthe Laplace equation are given by Eqs. (2.38–2.40). Of these three solutions, we have tochoose that solution which is consistent with the physical nature of the problem and the givenboundary conditions as depicted in Fig. 2.3.

x

y

O

y b = u = 0

x a = x = 0u = 0u = 0

y = 0u f x = ( )

Fig. 2.3 Dirichlet boundary conditions.

Consider the solution given by Eq. (2.38):

1 2 3 4( , ) ( ) ( cos sin )px pxu x y c e c e c py c py��

Using the boundary condition: (0, ) 0,u y � we get

1 2 3 4( ) ( cos sin ) 0c c c py c py �

which means that either 1 2 0c c� � or 3 4cos sin 0.c py c py� � But 3 4cos sin 0;c py c py� therefore,

1 2 0c c� � (2.42)

Again, using the BC; ( , ) 0,u a y � Eq. (2.38) gives

1 2 3 4( ) ( cos sin ) 0ap apc e c e c py c py��

implying thereby

1 2 0ap apc e c e�� (2.43)

To determine the constants c1, c2, we have to solve Eqs. (2.42) and (2.43); being homogeneous,the determinant



1 1 0ap ape e�

�

for the existence of non-trivial solution, which is not the case. Hence, only the trivial solution

( , ) 0u x y � is possible.

If we consider the solution given by Eq. (2.39) 5 6 7 8( , ) ( ) ( ),u x y c x c c y c� the boundary

conditions: (0, ) ( , ) 0u y u a y� � again yield a trivial solution. Hence, the possible solutions

given by Eqs. (2.38) and (2.39) are ruled out. Therefore, the only possible solution obtainedfrom Eq. (2.40) is

9 10 11 12( , ) ( cos sin ) ( )py pyu x y c px c px c e c e��

Using the BC: (0, ) 0,u y � we get 9 0.c � Also, the other BC: ( , ) 0u a y � yields

10 11 12sin ( ) 0py pyc pa c e c e��

For non-trivial solution, c10 cannot be zero, implying sin 0,pa � which is possible if pa nπ� or

/ , 1, 2, 3,p n a n� � �π Therefore, the possible non-trivial solution after using the superposition

principle is

1

( , ) sin [ exp ( / ) exp ( / )]n nn

n xu x y a n y a b n y a

a

π π π�

�� (2.44)

Now, using the BC: ( , ) 0,u x b � we get

sin [ exp ( / ) exp ( / )] 0n nn x

a n b a b n b aa

π π π� � �

implying thereby

exp ( / ) exp ( / ) 0n na n b a b n b aπ π � �

which gives

exp ( / ),

exp ( / )n nn b a

b an b a

ππ

� ��

1, 2, ,n � � �

The solution (2.44) now becomes

1

1

2 sin ( / ) exp { ( )/ } exp { ( )/ }( , )

exp ( / ) 2

2sin ( / )sin { ( )/ }

exp ( / )

n

n

n

n

a n x a n y b a n y b au x y

n b a

an x a h n y b a

n b a

π π ππ

π ππ

�

�

�

�

� � � ��

� ��

�

�



Let 2 /[exp ( / )] .n na n b a Aπ� � Then the solution can be written in the form

1

( , ) sin ( / ) sinh { ( )/ }nn

u x y A n x a n y b aπ π�

�� (2.45)

Finally, using the non-homogeneous boundary condition: ( , 0) ( ),u x f x� we get

1

sin ( / ) sinh ( / ) ( )nn

A n x a n b a f xπ π�

��

which is a half-range Fourier series. Therefore,

0

2sinh ( / ) ( ) sin ( / )

a

nA n b a f x n x a dxa

π π� � � (2.46)

Thus, the required solution for the given Dirichlet problem is

1

( , ) sin ( / ) sinh { ( )/ }nn

u x y A n x a n y b aπ π�

��

where

0

2 1( ) sin ( / )

sinh ( / )

a

nA f x n x a dxa n b a

ππ

��

��5 ��(��%�(�� 4�

The Neumann problem for a rectangle is defined as follows:

2PDE: 0, 0 , 0

BCs: (0, ) ( , ) 0, ( , 0) 0, ( , ) ( )x x y y

u x a y b

u y u a y u x u x b f x

� � � � � �

� � � � (2.48)

The general solution of the Laplace equation using the method of variables separable isgiven in Section 2.5, and is found to be

1 2 3 4( , ) ( cos sin ) ( )py pyu x y c px c px c e c e��

The BC: (0, ) 0xu y � gives

2 3 40 ( )py pyc p c e c e��

implying 2 0.c � Therefore,

1 3 4( , ) cos ( )py pyu x y c px c e c e�� (2.49)



The BC: ( , ) 0xu a y � gives

1 3 40 sin ( )py pyc p pa c e c e��

For non-trivial solution, 1 0,c implying

sin 0,pa � ,pa nπ� np

a

π� ( 0,1, 2, )n � �

Thus the possible solution is

/ /( , ) cos ( )n y a n y an xu x y Ae Be

aπ ππ �� (2.50)

Now, using the BC: ( , 0) 0,yu x � we get

0 cosn x n n

A Ba a a

π π π� ��

implying .B A� Thus, the solution is

( , ) cos [exp ( / ) exp ( / )]

2 cos cosh

n xu x y A n y a n y a

a

n x n yA

a a

π π π

π π

� � �

�

Using the superposition principle and defining 2 ,nA A� we get

0

cos coshnn

n x n yu A

a a

π π�

�� (2.51)

Finally, using the BC: ( , ) ( ),yu x b f x� we get

1

( ) cos sinhnn

n x n n bf x A

a a a

π π π�

��

which is the half-range Fourier cosine series. Therefore,

0

2sinh ( ) cos

a

nn n b n x

A f x dxa a a a

π π π� �Hence, the required solution is

01

cos coshnn

n x n yu A A

a a

π π�

�� (2.52)



where

0

2 1( ) cos

sinh ( / )

a

nn x

A f x dxn n b a a

ππ π

� �

��6 � �� %�(��

The Dirichlet problem for the circle is defined as follows:

2PDE: 0, 0 , 0 2

BC: ( , ) ( ), 0 2

u r a

u a f

θ π

θ θ θ π

� � � � � �

� � � (2.53)

where ( )f θ is a continuous function on IR .� The task is to find the value of u at any point

in the interior of the circle IR in terms of its values on IR� such that u is single valued and

continuous on IR.In view of circular geometry, it is natural to choose polar coordinates to solve this

problem and then use the variables separable method. The requirement of single-valuedness

of u in IR implies the periodicity condition, i.e.,

( , 2 ) ( , ),u r u rθ π θ� � 0 ,r a� � (2.54)

From Eq. (2.27), 2 0u� � which in polar coordinates can be written as

2

1 10rr ru u u

r rθθ� � �

If ( , ) ( ) ( ),u r R r Hθ θ� the above equation reduces to

2

1 10R H R H RH

r r� � ��

This equation can be rewritten as

2r R rR Hk

R H

�� (2.55)

which means that a function of r is equal to a function of θ and, therefore, each must beequal to a constant k (a separation constant).

Case I Let 2.k λ� Then

2 2 0r R rR Rλ � � (2.56)

which is a Euler type of equation and can be solved by setting .zr e� Its solution is

1 2 1 2z zR c e c e c r c rλ λ λ λ� ��



Also,2 0H H � λ

whose solution is

3 4cos sinH c cλθ λθ� �Therefore,

1 2 3 4( , ) ( ) ( cos sin )u r c r c r c cλ λθ λθ λθ�� (2.57)

Case II Let 2.k λ� � Then

2 2 0,r R rR R� � �� λ 2 0H H� � λTheir respective solutions are

1 2

3 4

cos ( ln ) sin ( ln )R c r c r

H c e c eλθ λθ

λ λ�

� �

� �

Thus

1 2 3 4( , ) [ cos ( ln ) sin ( ln )] ( )u r c r c r c e c eλθ λθθ λ λ �� (2.58)

Case III Let 0.k � Then we have

0rR R� ��

Setting ( ) ( ),R r V r�� we obtain

0,dV

r Vdr

� � �� 0dV dr

V r� �

Integrating, we get 1ln ln .Vr c� Therefore,

1c dRV

r dr� �

On integration,

1 2lnR c r c�

Also,

0H ��After integrating twice, we get

3 4H c cθ� �

Thus,

1 2 3 4( , ) ( ln ) ( )u r c r c c cθ θ� � � (2.59)



Now, for the interior problem, 0r � is a point in the domain IR and since ln r is not defined

at 0,r � the solutions (2.58) and (2.59) are not acceptable. Thus the required solution is

obtained from Eq. (2.57). The periodicity condition in θ implies

3 4 3 4cos sin cos ( ( 2 )) sin ( ( 2 ))c c c cλθ λθ λ θ π λ θ π �

i.e.

3 4[cos cos ( 2 )] [sin sin ( 2 )] 0c cλθ λθ λπ λθ λθ λπ� � � � � �

or

3 42 sin [ sin ( ) cos ( )] 0c c � �λπ λθ λπ λθ λπ

implying sin 0, , ( 0,1, 2, ).n n nλπ λπ π λ� � � � � Using the principle of superposition and

renaming the constants, the acceptable general solution can be written as

0

( , ) ( ) ( cos sin )n nn n n n

n

u r c r d r a n b nθ θ θ�

�

�� (2.60)

At r = 0, the solution should be finite, which requires 0.nd � Thus the appropriate solution

assumes the form

0

( , ) ( cos sin )nn n

n

u r r A n B nθ θ θ�

��

For 0,n � let the constant A0 be A0/2. Then the solution is

0

1

( , ) ( cos sin )2

nn n

n

Au r r A n B nθ θ θ

�

�� (2.61)

which is a full-range Fourier series. Now we have to determine An and Bn so that the BC:

( , ) ( )u a fθ θ� is satisfied, i.e.,

0

( ) ( cos sin )nn n

n

f a A n B nθ θ θ�

��

Hence,

2

00

2

0

2

0

1( )

1( ) cos

1( ) sin , 1, 2, 3,

nn

nn

A f d

a A f n d

a B f n d n

π

π

π

θ θπ

θ θ θπ

θ θ θπ

�

�

� � �

�

�

�

(2.62)



In Eqs. (2.62) we replace the dummy variable θ by φ to distinguish this variable from the

current variable θ in Eq. (2.61). Substituting Eq. (2.62) into Eq. (2.61), we obtain the relation

2 2

0 01

2

0

1 cos( , ) ( ) cos ( ) ( )

2

sinsin ( ) ( )

n

nn

n

n

r nu r f d n f d

a

r nn f d

a

π π

π

θθ φ φ φ φ φπ π

θ φ φ φπ

�

�

��

��

��

��

�Interchanging the order of summation and integration, we get

2 2

0 01

2

01

1 1( , ) ( ) ( ) {cos cos sin sin }

2

1 1( ) cos ( )

2

n

n

n

n

ru r f d f n n n n d

a

rf n d

a

π π

π

θ φ φ φ φ θ φ θ φπ π

φ φ θ φπ

�

�

�

�

� ��

� ��

��

�� (2.63)

To obtain an alternative expression for ( , )u r θ in closed integral form, we can proceed as

follows:Let

1

1

cos ( )

sin ( )

n

n

n

n

rc n

a

rs n

a

�

�

�

�

� � ��

� � ��

�

�

φ θ

φ θ

so that

( )

1

ni

n

rc is e

aφ θ

��

�

� � � ��

Since , ( / ) 1r a r a� � and ( )| | 1,ie φ θ� �

( )( )

( )1

( )

( ) ( )

( / )

[1 ( / ) ]

( / ){ ( / )}

[1 ( / ) ][1 ( / ) ]

n ii

in

i

i i

r r a ec is e

a r a e

r a e r a

r a e r a e

φ θφ θ

φ θ

φ θ

φ θ φ θ

�

�

� ��

�

�



Equating the real part on both sides, we get

2 2

2 2

[ / ) cos ( ) ( / )]

[1 (2 / ) cos ( ) ( / )]

r a r ac

r a r a

φ θφ θ� ��

� � �

Thus, the expression in the square brackets of Eq. (2.63) becomes

2 2 2 2

2 2 2 2

1 [( / ) cos ( ) ( / )]

2 [1 (2 / ) cos ( ) ( / )] 2[ 2 cos ( ) ]

r a r a a r

r a r a a ar r

φ θφ θ φ θ� � ��

� � � � � �

Thus, the required solution takes the form

2 22

2 20

1 ( ) ( )( , )

2 [ 2 cos ( ) ]

a r fu r d

a ar r

π φθ φπ φ θ

�� (2.64)

This is known as Poisson’s integral formula for a circle, which gives a unique solution forthe Dirichlet problem. The solution (2.64) can be interpreted physically in many ways: It can

be thought of as finding the potential ( , )u r θ as a weighted average of the boundary

potentials ( )f φ weighted by the Poisson kernel P, given by

2 2

2 2[ 2 cos ( ) ]

a rP

a ar rφ θ��

� � �

It can also be thought of as a steady temperature distribution ( , )u r θ in a circular disc, when

the temperature u on its boundary IR� is given by ( )u f φ� which is independent of time.

��7 8 �� %�(��

The exterior Dirichlet problem is described by

2PDE: 0

BC: ( , ) ( )

u

u a f

� �

�θ θ (2.65)

u must be bounded as .r ��

By the method of separation of variables, the general solution (2.60) of 2 0u� � in polarcoordinates can be written as

0

( , ) ( ) ( cos sin )n nn n n n

n


�

��

Now as, ,r �� we require u to be bounded, and, therefore, 0.nc �



After adjusting the constants, the general solution now reads

0( , ) ( cos sin )n

n nn

u r r A n B nθ θ θ∞

−

== +∑

With no loss of generality, it can also be written as

0

1( , ) ( cos sin )

2n

n nn

Au r r A n B nθ θ θ

∞−

== + +∑ (2.66)

Using the BC: ( , ) ( ),u a fθ θ= we obtain

0

1( ) ( cos sin )

2n

n nn

Af a A n B nθ θ θ

∞−

== + +∑

This is a full-range Fourier series in ( ),f θ where

20 0

2

0

2

0

1 ( )

1 ( ) cos

1 ( ) sin

nn

nn

A f d

a A f n d

a B f n d

−

−

=

=

=

∫

∫

∫

π

π

π

θ θπ

θ θ θπ

θ θ θπ

(2.67)

In Eq. (2.67) we replace the dummy variable θ by φ so as to distinguish it from the current

variable .θ We then introduce the changed variable into solution (2.66) which becomes

2 2

0 01

2

0

1( , ) ( ) cos cos ( ) ( )2

sin sin ( ) ( )

n n

n

n n

r au r f d n n f d

r a n n f d

∞ −

=

−

⎡= + ⎢

⎢⎣

⎤+ ⎥

⎥⎦

∑∫ ∫

∫

π π

π

θ φ φ θ φ φ φπ π

θ φ φ φπ

or

2

0 1

1 1( , ) ( ) cos ( )2

n

n

au r f n dr

πθ φ φ θ φ

π

∞

=

⎡ ⎤⎛ ⎞⎢ ⎥= + −⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦∑∫ (2.68)

Let

1

cos ( )n

n

aC n

rf q

•

=

Ê ˆ= -Á ˜Ë ¯Â



�1

sin ( )n

n

aS n

r� �

�

�

� �� Then,

( )

1

ni

n

aC iS e

rφ θ

��

�

� ��

Since ( )1, | | 1.iae

rφ θ�� We have

( ) )

( ) ( ) ( )

( / ) [ ( / )]

[1 ( / ) ] [1 ( / ) ] [1 ( / ) ]

i i

i i i

a e a r e a rC iS

r a r e a r e a r e

φ θ φ θ

φ θ φ θ φ θ

� � �

� � � ��

� � ��

Hence,

2 2

2 2

[( / ) cos ( ) ( / )]

[1 (2 / ) cos ( ) ( / )]

a r a rC

a r a r

φ θφ θ� ��

� � �

Thus the quantity in the square brackets on the right-hand side of Eq. (2.68) becomes

2 2 2 2

2 2 2 2

1 [( / ) cos ( ) ( / )]

2 [1 (2 / ) cos ( ) ( / )] 2[ 2 cos ( ) ]

a r a r r a

a r a r r ar a

φ θφ θ φ θ� � ��

� � � � � �

Therefore, the solution of the exterior Dirichlet problem reduces to that of an integral equationof the form

2 22

2 20

1 ( ) ( )( , )

2 [ 2 cos ( ) ]

r a fu r d

r ar a

π φθ φπ φ θ

�� (2.69)

EXAMPLE 2.4 Find the steady state temperature distribution in a semi-circular plate ofradius a, insulated on both the faces with its curved boundary kept at a constant temperatureU0 and its bounding diameter kept at zero temperature as described in Fig. 2.4.

r

� � = /2

u u = 0

r a =

� � = � = 0u = 0

�

Fig. 2.4 Semi-circular plate.



Solution The governing heat flow equation is

2tu u= ∇

In the steady state, the temperature is independent of time; hence 0,tu = and the temperature

satisfies the Laplace equation. The problem can now be stated as follows: To solve

22

0

1 1PDE: ( , ) 0

BCs: ( , ) , ( , 0) 0, ( , ) 0

rr ru r u u ur r

u a U u r u r

∇ = + + =

= = =

θθθ

θ π

the acceptable general solution is

( , ) ( ) ( cos sin )u r cr Dr A Bλ λθ λθ λθ−= + + (2.70)

From the BC: ( , ) 0,u r 0 = we get 0;A = however, the BC: ( , ) 0u r π = also gives

sin ( ) 0B cr Drλ λλπ −+ =

implying either 0B = or sin 0. 0Bλπ = = gives a trivial solution. For a non-trivial solution,

we must have sin 0,λπ = implying

,nλπ π= 1, 2,n = …

meaning thereby .nλ = Hence, the possible solution is

( , ) sin ( )u r B n Cr Drλ λθ θ −= + (2.71)

In Eq. (2.71), we observe that as 0,r → the term .r λ− → ∞ But the solution should be finite

at 0,r = and so 0.D = Then after adjusting the constants, it follows from the superpositionprinciple that,

1( , ) sinn

nn

u r B r nθ θ∞

== ∑

Finally, using the first BC: 0( , ) ,u a Uθ = we get

01

( , ) sinnn

nu a U B a nθ θ

∞

== = ∑



which is a half-range Fourier sine series. Therefore,

0

00

4, for 1, 3,2

sin

0, for 2, 4,

nn

Un

nB a U n d

n

π πθ θπ

� � ��

�

Hence,

04,n n

UB

n aπ� 1, 3,n � �

With these values of Bn, the required solution is

0

odd

4 1( , ) sin

n

n

U ru r n

n aθ θ

π

�

�

��

��9 � ��(��%�(��

The interior Neumann problem for a circle is described by

2PDE: 0, 0 ; 0 2

( , )BC: ( ),

u r a

u u ag r a

n r

� ��

� � �

� � �

θ π

θ θ

(2.72)

Following the method of separation of variables, the general solution (2.60) of equation2 0u� � in polar coordinates is given by

0

( , ) ( ) ( cos sin )n nn n n n

n


�

��

At r = 0, the solution should be finite and, therefore, 0.nd � Hence, after adjusting the constants,

the general solution becomes

0


n

u r r A n B nθ θ θ�

��

With no loss of generality, this equation can be written as

0

1

1

1

( , ) ( cos sin )2

( cos sin )

nn n

n

nn n

n

Au r r A n B n

unr A n B n

r

��

�

�

�

�

� � �

� �

�

�

θ θ θ

θ θ

��



Using the BC:

( , ) ( )u

a gr

��

�θ θ

we get

1

1

( ) ( cos sin )nn n

n

g na A n B nθ θ θ�

�

�� (2.74)

which is a full-range Fourier series in ( ),g θ where

21

0

1

0

1( ) cos

1( ) sin

nn

nn

na A g n d

na B g n d

π

π

θ θ θπ

θ θ θπ

�

��

�

�

�

� (2.75)

Here, we replace the dummy variable θ by φ to distinguish from the current variable θ in

Eq. (2.74). Now introducing Eq. (2.75) into Eq. (2.73), we obtain

201 0

1

( , ) ( ) (cos cos sin sin )2

n

nn

A ru r g n n n n d

n a

πθ φ φ θ φ θ φ

π

�

��

� � �or

20

01

( , ) ( ) cos ( )2

n

n

A r au r g n d

a n

πθ φ φ θ φ

π

�

�

� �� (2.76)

This solution can also be expressed in an alternative integral form as follows: Let

cos ( )

sin ( )

n

n

r aC n

a n

r aS n

a n

φ θπ

φ θπ

� � � ��

� � � ��

!

!Therefore,

( ) ( )

1

2 3( ) ( ) ( )

1

1 2 3

n nin i

n

i i i

r a a rC iS e e

a n a n

r r re e e

a a a a

φ θ φ θ

φ θ φ θ φ θ

π π

π

��

�

� � �

� � � � � � � � � � � �

� ��

� �

�



or

( )ln 1 ln 1 cos ( ) sin ( )ia r a r rC iS e ia a a

φ θ φ θ φ θπ π

−⎡ ⎤ ⎡ ⎤+ = − − = − − − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦(2.77)

To get the real part of ln z, we may note that

ln or WW z z e= =

i.e., cos sin .u iv u ux iy e e v ie v++ = = + Therefore,

2 2 2 2

cos , sin

| |

u u

u

x e v y e v

e x y z

= =

= + =

i.e., ln | | .u z= Therefore,

2 2

2 2

2

ln 1 cos ( ) sin ( )

2 cos ( )ln

a r rCa a

a a ar ra

φ θ φ θπ

φ θπ

⎛ ⎞ ⎛ ⎞= − − − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− − += −

Thus the required solution is

2 22020

2 cos ( )( , ) ln ( )2A a a ar ru r g d

a

π φ θθ φ φπ

− − += − ∫ (2.78)

which is again an integral equation.

2.11 SOLUTION OF LAPLACE EQUATION IN CYLINDRICAL COORDINATES

The Laplace equation in cylindrical coordinates assumes the following form:

22

1 1 0rr r zzu u u u ur r θθ∇ = + + + = (2.79)

We now seek a separable solution of the form

( , , ) ( , ) ( )u r z F r Z zθ θ= (2.80)

Substituting Eq. (2.80) into Eq. (2.79), we get

2 2 2

2 2 2 21 1 0F F F d ZZ Z Z Fr rr r dzθ

+ + + =∂ ∂ ∂∂∂ ∂

or

2 2 2

2 2 2 21 1 1 1 (say)F F F d Z kr r F Zr r dzθ

⎛ ⎞+ + = − =⎜ ⎟⎝ ⎠

∂ ∂ ∂∂∂ ∂



where k is a separation constant. Therefore, either

2

20

d ZkZ

dz� � (2.81)

or

2 2

2 2 2

1 10

F F FKF

r rr r

� � ��

� � � �θ

(2.82)

If k is real and positive, the solution of Eq. (2.81) is

1 2cos sinZ c k z c k z� �

If k is negative, the solution of Eq. (2.81) is

1 2k z k zZ c e c e��

If k is equal to zero, the solution of Eq. (2.81) is

1 2Z c z c� From physical considerations, one would expect a solution which decays with increasing z

and, therefore, the solution corresponding to negative k is acceptable. Let 2.k λ� � Then

1 2z zZ c e c eλ λ�� (2.83)

Equation (2.82) now becomes

2 22

2 2 2

1 10

F F FF

r rr r

� � ��

� � � �λθ

Let ( , ) ( ) ( ).F r f r Hθ θ� Substituting into the above equations, we get

22

1 10f H f H f H f H

r r� � � �� λ

or

2 2 2 1( ) (say)

Hr f r f r f k

f H

�� λ

From physical consideration, we expect the solution to be periodic in ,θ which can be obtained

when k � is positive and 2.k n� Therefore, the acceptable solution will be

3 4cos sinH c n c nθ θ� (2.84)

When 2,k n�� we will also have

22 2 2 2

2( ) 0

d f dfr r r n f

drdrλ� � � � (2.85)



which is a Bessel’s equation whose general solution is given by

( ) ( )n nf AJ r BY rλ λ� (2.86)

Here, ( )nJ rλ and ( )nY rλ are the nth order Bessel functions of first and second kind, respectively.

Since ( ) as 0, ( )n nY r r Y rλ λ� �� becomes unbounded at 0.r � Continuity of the solution

demands 0.B � Hence the most general and acceptable solution of 2 0u� � is

1 2 3 4( , , ) ( ) ( ) ( cos sin )� �� z znu r z J r c e c e c n c n (2.87)

EXAMPLE 2.5 A homogeneous thermally conducting cylinder occupies the region 0 ,r a� �0 2 , 0 ,z hθ π where , ,r zθ are cylindrical coordinates. The top z h� and the lateral

surface r a� are held at 0°, while the base 0z � is held at 100°. Assuming that there are nosources of heat generation within the cylinder, find the steady-temperature distribution withinthe cylinder.

Solution The temperature u must be a single valued continuous function. The steadystate temperature satisfies the Laplace equation inside the cylinder. To compute the temperaturedistribution inside the cylinder, we have to solve the following BVP:

2PDE: 0

BCs: 0 on ,

0 on ,

100 on 0

u

u z h

u r a

u z

� �

� � �

� � �

� � �The general solution of the Laplace equation in cylindrical coordinates as given in Section 2.11is

1 2 3 4( , , ) ( ) ( cos sin ) ( )z znr r z J r c n c n c e c eλ λθ λ θ θ ��

Since the face 0z � is maintained at 100° and since the other face and lateral surface of thecylinder are maintained at 0°, the temperature at any point inside the cylinder is obviously

independent of .θ This is possible only when 0n � in the general solution. Thus,

0( , ) ( ) ( )z zu r z J r Ae Beλ λλ ��

Using the BC: 0u � on ,z h� we get

00 ( ) ( )h hJ r Ae Beλ λλ ��

implying thereby 0,h hAe Beλ λ�� from which

h

h

AeB

e

λ

λ��



Therefore, the solution is

( ) ( )0 ( )( , ) [ ]z h z h

h

J r Au r z e e

eλ λ

λλ � � ��

or

0 1( , ) ( ) sinh ( )u r z J r A z hλ λ� �

where 1 2 / .hA A e λ�� Now using the BC: 0 on ,u r a� � we have

1 00 ( ) sinh ( )A J a z hλ λ� �

implying 0 ( ) 0,J aλ � which has infinitely many positive roots. Denoting them by ,nξ we

have ,n aξ λ� and therefore,

n

a

ξλ �

Thus the solution is

1 0( , ) sinh ( ) ,n nru r z A J z h

a a

ξ ξ� � � �� 1, 2,n � �

Using the principle of superposition, we have

01

( , ) sinh ( )n nn

n

ru r z A J z h

a a

ξ ξ�

�

� � �� The BC: u = 100° on z = 0 gives

01

100 sinh n nn

n

h rA J

a a

ξ ξ�

�

� ��

which is a Fourier-Bessel series. Multiplying both sides with 0 ( / )mrJ r aξ and integrating, we

get

0 0 00 0

1

100 sinha a

m n n mn

n

r h r rrJ dr A rJ J dr

a a a a

ξ ξ ξ ξ�

�

� � � � � � � �� Using the orthogonality property of Bessel’s function, namely,

220

1

0, if

( ) ( )( ), if

2

a

n i n jin

i j

xJ x J x dx aJ i j

α αα�

��

��



where ,i jα α are the zeros at ( ) 0,nJ x � we have

22

0 101

100 sinh ( )2

an n

n nn

r h arJ dr A J

a a

ξ ξ ξ�

�

� �� Therefore,

0022

1

200

sinh ( )

an

nn

n

rA rJ dr

h aa J

a

ξξ ξ

� � ��

Setting

,nrx

a

ξ�

n

adr dx

ξ�

the relation for An can also be written as

0022

1

200( )

sinh ( )

nn

nn n

A xJ x dxh

Ja

ξ

ξξ ξ�

� ��

Using the recurrence relation

1( ) [ ( )],n nn n

dx J x x J x

dx� �

For 1,n � we get

0 1( ) ( )xJ x dx xJ x��Now, An can be written as

122

11 0

200 ( ) 200

sinh ( / ) ( )sinh ( / ) ( )

n

nn n nn n n

x J xA

h a Jh a J

ξ

ξ ξ ξξ ξ ξ

� ��

��

Hence, the required temperature distribution inside the cylinder is

0

11

( / ) sinh [( / ) ( )]( , ) 200

sinh ( / ) ( )n n

n n nn

J r a a z hu r z

h a J

ξ ξξ ξ ξ

�

�

��

��

where nξ are the positive zeros of 0 ( ).J ξ



EXAMPLE 2.6 Find the potential u inside the cylinder 0 , 0 2 , 0 ,r a z hθ π� � � � � � if

the potential on the top ,z h� and on the lateral surface r = a is held at zero, while on the

base z = 0, the potential is given by 2 20( , , 0) (1 / ),u r V r aθ � � where V0 is a constant; , ,r zθ are

cylindrical polar coordinates.

Solution The potential u must be a single-valued continuous function and satisfy theLaplace equation inside the cylinder. To compute the potential inside the cylinder, we haveto solve the following BVP:

2

2

0 2

PDE: 0

BCs: 0 on ,

0 on ,

1 on 0

u

u z h

u r a

ru V z

a

� �

� �

� �

� ��

In cylindrical coordinates, the general solution of the Laplace equation as given in Section 2.11is

1 2 3 4( , , ) ( ) ( cos sin ) ( )z znu r z J r c n c n c e c eλ λθ λ θ θ ��

Since the face z = 0 has potential 2 20 (1 / ),V r a� which is purely a function of r and is independent

of θ and since the other faces of the cylinder are at zero potential, the potential at any point

inside the cylinder will obviously be independent of .θ This is possible only when n = 0 inthe general solution. Thus,

0( , ) ( ) ( )z zu r z J r Ae Beλ λλ ��

Using the BC: u = 0 on z = h, we obtain

00 ( ) ( )h hJ r Ae Beλ λλ ��

implying 0,h hAe Beλ λ�� which yields

h

h

AeB

e

λ

λ��

Hence, the solution is

( ) ( )0( , ) ( )[ ]z h z h

h

Au r z J r e e

eλ λ

λ λ � � ��



or

1 0( , ) ( ) sinh ( )u r z A J r z hλ λ� �

where 1 / .hA A e λ�� Now, using the BC: u = 0 on the lateral surface, i.e., on r = a, we get

1 00 ( ) sinh ( )A J a z hλ λ� �

implying 0 ( ) 0.J aλ � This has infinitely many positive roots; denoting them by nξ we shall

have

n aξ λ� or /n aλ ξ�

The solution now takes the form

1 0( , ) sinh ( ) ,n nru r z A J z h

a a

ξ ξ� � � �� 1, 2,n � �

The principle of superposition gives

01

( , ) sinh ( )n nn

n

ru r z A J z h

a a

ξ ξ�

�

� � � ��

The last BC:2

0 21

ru V

a

� ��

on 0z � yields

2

0 021

1 sinh n nn

n

h rrV A J

a aa

ξ ξ�

�

� � � � � ��

This is a Fourier-Bessel’s series. Multiplying both sides by 0 ( / )mrJ r aξ and integrating, we get

2

0 0 0 020 01

1 sinha a

m n n mn

n

r h r rrV rJ dr A rJ J dr

a a a aa

ξ ξ ξ ξ�

�

� � � � � � � � � �� Using the orthogonality property of the Bessel functions

220

1

0, if

( ) ( )( ), if

2

� ��

��

��

a

n i n j

n i

i j

xJ x J x dx aJ i j

where ,i jα α are the zeros of ( ) 0,nJ x � we get

2 22

0 0 1201

1 sinh ( )2

an n

n nn

r hr aV rJ dr A J

a aa

ξ ξ ξ�

�

� � � � � ��



which gives

20

020221

21

sinh ( )

an

nn

n

V rrA rJ dr

h aaa Ja

ξξ ξ

� � � ��

By letting / ,nr a xξ � this equation can be modified to

2 200

0241

2( ) ( )

sinh ( )

nn n

nn n

VA x xJ x dx

hJ

a

ξξ

ξξ ξ� �

� ��

Using the well-known recurrence relation

1( ) [ ( )]adx J x x J x

dxα

α α� � for 1, 2,� � �

we get

0 1( ) ( ),xJ x xJ x�� 2 21 2( ) ( )x J x x J x��

From these relations, we obtain

2 201

0241

2( ) [ ( )]

sinh ( )

nn n

nn n

VA x d xJ x

hJ

a

ξξ

ξξ ξ� �

� ��

Integrating by parts, we get

20124 0

1

20224 0

1

202 024

1

4( )

sinh ( / ) ( )

4[ ( )]

sinh ( / ) ( )

4[ ( )]

sinh ( / ) ( )

n

n

n

nn n n

n n n

n n n

VA x J x dx

h a J

Vd x J x

h a J

Vx J x

h a J

ξ

ξ

ξ

ξ ξ ξ

ξ ξ ξ

ξ ξ ξ

��

��

��

Thus,

0 2221

4 ( )

sinh ( / ) ( )n

nn n n

V JA

h a J

ξξ ξ ξ

��



The recurrence relation

1 12

( ) ( ) ( )n n nn

J x J x J xx� ��

for 1n � gives

0 2 12

( ) ( ) ( )n n nn

J J Jξ ξ ξξ

� �

Hence,

2 12

( ) ( )n nn

J Jξ ξξ

�

since 0 ( ) 0.nJ ξ � Therefore,

0 1231

8 ( )

sinh ( / ) ( )n

nn n n

V JA

h a J

ξξ ξ ξ

��

Thus, the required potential inside the cylinder is

0 0

311

8 sinh ( )( , )

( ) sinh

n n

nnn n

rV J z h

a au r z

hJ

a

ξ ξ

ξξ ξ

�

�

� � � � � � � � � ��

�

��

In Example 2.3, the Laplace equation is expressed in spherical coordinates and has thefollowing form:

22 2

2 2

1 1sin 0

sin sin

u u uu r

r r

� � � � ��

� � � �� θ

θ θ θ θ φ(2.88)

Let us assume the separable solution in the form

( , , ) ( ) ( , )u r R r Fθ φ θ φ� (2.89)


22

2 2sin 0

sin sin�

� � � � ��

R R F R FF r

r r

� � � � ��

Separation of variables gives

22

2

1 1sin

sin sinF Fd dR

rdr dr

R F

� � ��

� �� θ

θ θ θ θ φμ



where μ is a separation constant. Therefore,

21 d dRr

R dr drμ� � � �� (2.90)

2

2

1 1sin

sin sin

F F

Fθ μ

θ θ θ θ φ

� ��

� � ��

(2.91)

Equation (2.90) gives

22

22 0

d R dRr r R

drdrμ� � �

which is a Euler’s equation. Hence, using the transformation ,zr e� the auxiliary equationcan be written as

2( 1) 2 0D D D D Dμ μ� � � � � � �

where / .D d dz� Its roots are given by

1 1 4

2D

μ� � ��

Let ( 1);μ α α� � � then we get

211 2

2 1 1

2 2 2D

α

α

� ��

Hence, D α� and ( 1).α� � Therefore, the solution of Euler’s equation is

( 1)1 2R c r c rα α� �� (2.92)

Taking ( 1),μ α α� � � Eq. (2.91) becomes

2

2

1sin ( 1) sin 0

sin

F FF

� � ��

� � � � � �� θ α α θ

θ θ θ φInserting

( ) ( )F H θ φΦ�

into the above equation and separating the variables, we obtain

22

2

sin 1sin ( 1) sin

d dH dH

H d d d

θ θ α α θ νθ θ φ

ΦΦ

� ��



where v2 is another separation constant. Then

22

20

d

dν

φΦ Φ� � (2.93)

2sinsin ( 1) sin ( )

d dHH

H d d

θ θ α α θ νθ θ

� �� (2.94)

The general solution of Eq. (2.93) is

3 4cos sinc cνφ νφΦ � � (2.95)

provided 0.ν � If 0,ν � the solution is independent of φ which corresponds to the axisymmetriccase. Equation (2.94) becomes, for the axisymmetric, case,

sin ( 1) sin ( ) 0d dH

Hd d

� � � � �� θ α α θ

θ θ

Transforming the independent variable θ to x and by letting cos ,x � θ the abvoe equation

becomes

sin ( 1) sin ( ) 0d dH dx dx

Hdx dx d d

θ α α θθ θ

� � ��

i.e.,

2(1 cos ) ( 1) 0d dH

Hdx dx

θ α α� �� or

2(1 ) ( 1) 0d dH

x Hdx dx

α α� �� (2.96)

This is the well-known Legendre equation. Its general solution is given by

5 6( ) ( ),H c P x c Q xα α� � 1 1x� � � (2.97)

where ,P Qα α are Legendre functions of the first and second kind respectively. For convenience

let α be a positive integer, say .nα � Then

5 6(cos ) (cos )n nH c P c Qθ θ� � (2.98)

Continuity of ( )H θ at 0,θ π� implies the continuity of ( )H x at 1.x � � Since Qn(x) has a

singularity at x = 1, we choose 6 0.c � Therefore, in axisymmetric case the solution of Laplace

equation in spherical coordinates is given by

( 1)1 2 3 5( , , ) { } ( ) [ (cos )]nu r c r c r c c Pα αθ φ θ� ��



After renaming the constants and using the principle of superposition, we find the solutionto be

( 1)

0

( , ) [ ] (cos )n nn n n

n

u r A r B r Pθ θ�

� �

�� (2.99)

EXAMPLE 2.7 In a solid sphere of radius ‘a’, the surface is maintained at the temperaturegiven by

cos , 0 /2( )

0, /2

kf

� ��

θ θ πθ

π θ π

Prove that the steady state temperature within the solid is

2 4

0 1 2 41 1 5 3

( , ) (cos ) (cos ) (cos ) (cos )4 2 16 32

r r ru r k P P P P

a a a

� �� θ θ θ θ θ

Solution It is known that the steady state temperature distribution is governed bythe Laplace equation. In spherical polar coordinates, the axisymmetric solution of the Laplaceequation in general with the assumption that the temperature should be finite at the origin isgiven by Eq. (2.99) in the form

0

( , ) (cos )nn n

n

u r A r Pθ θ�

�� (2.100)

Using the given BC: ( , ) ( ),u a fθ θ� we have

0 0

( , ) ( ) (cos ) (cos )nn n n n

n n

u a f A a P b Pθ θ θ θ� �

� ��

where .nn nA a b� This is a Fourier-Legendre series, where

1

1

2 1( ) (cos )

2n nn

b f P dθ θ θ�

��

In the present problem,

1

0

2 1( ) (cos )

2n nn

b f P dθ θ θ�� Let cos ,xθ �

1 1

0 00 0

1 1( ) 1

2 2 4

kb kx P x dx kx dx� � � � � ��



Hence, we get

0 4kA =

Also,

11 10

32 2

kb kx x dx A a= ⋅ ⋅ = =∫Therefore,

1

21 12 20 0

12

5 5 3 1 5( )2 2 2 16

kAa

xb kxP x dx kx dx k

⎛ ⎞= ⎜ ⎟⎝ ⎠

−= = =∫ ∫Thus,

2 25 1

16A k

a= ⋅

Further,

31 13 30 0

7 7 5 3( ) 02 2 2

x xb kxP x dx kx dx−= = =∫ ∫

Similarly, noting that 4 24

1( ) (35 30 3),8

P x x x= − + we get

44 4

332

b k A a= − =

Hence,

4 43 1 ,32

A ka

= − ⋅ L

Substituting these values of A0, A1, A2,… into Eq. (2.100), we obtain, finally, the requiredtemperature as

0 1

2 4

2 4

1 1( , ) (cos ) (cos )4 2

5 3(cos ) (cos )16 32

ru r k P Pa

r rP Pa a

θ θ θ

θ θ

⎡ ⎛ ⎞= + ⎜ ⎟⎢ ⎝ ⎠⎣

⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − + ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎥⎦L .



EXAMPLE 2.8 Find the potential at all points of space inside and outside of a sphere ofradius R = 1 which is maintained at a constant distribution of electric potential

( , ) ( ) cos 2 .u R fθ θ θ� �

Solution It is known that the potential on the surface of a sphere is governed by theLaplace equation. The Laplace equation in spherical polar coordinates is

2 2 2

2 2 2 2 2 2 2

2 1 cot 10

sin

u u u u u

r rr r r r

� � � � ��

� � � � �θθθ θ φ

The possible general solution by variables separable method, after using superposition principle,is given by Eq. (2.99). Thus we have two possible solutions:

10

2 10

( , ) (cos )

( , ) (cos )

nn n

n

nnn

n

u r A r P

Bu r P

r

θ θ

θ θ

�

�

�

��

�

�

�

�

(2.101)

(2.102)

For points inside the sphere, we take the series (2.101). Why is this so? Applying the BC: ( , )u R θ

( ) cos 2 ,f θ θ� � we obtain

0

( ) (cos )nn n

n

f A R Pθ θ�

��

which is a generalized Fourier series of ( )f θ in terms of the Legendre polynomials. Using

the orthogonality property, we get

1

1

2 1( ) ( )

2n

n nn

A R f P x dxθ�

�� Let cos .x θ� Then we have

0

2 1( ) (cos ) sin

2n nn

nA f P d

R

πθ θ θ θ��

For points outside the sphere, we take the series (2.102). Why is this so? Using the BC:

( , )u R θ ( ),f θ� we get

10

( ) (cos )nnn

n

Bf P

Rθ θ

�

�

� �Again, using the orthogonality property of Legendre polynomials, we have

1

0

2 1( ) (cos )sin

2n

n nn

B R f P dπ

θ θ θ θ��



In the present problem, it is assumed that at 2 21, ( ) cos 2 2 cos 1 2 1.R f xθ θ θ� � � � � � Hence,

1 2

1

2 1(2 1) ( )

2n nn

A x P x dx�

�� However,

22

1( ) (3 1)

2P x x� �

Therefore,

22

4 12 1 ( )

3 3x P x� � �

Thus,

1 1

2 01 1

2 1 4 1( ) ( ) ( ) ( )

2 3 3n n nn

A P x P x dx P x P x dx� �

� � �� Using the orthogonality property of Legendre polynomials, all integrals vanish except thosecorresponding to n = 0 and n = 2. We obtain, therefore,

1 20 01

1 22 21

1 1 1( )

2 3 3

5 4 4( )

2 3 3

A P x dx

A P x dx

�

�

� � � � �

� � �

�

�Also,

1 2

1

1 1

2 01 1

2 1(2 1) ( )

2

2 1 4 1( ) ( ) ( ) ( )

2 3 3

n n

n n

nB x P x dx

nP x P x dx P x P x dx

�

� �

��

� � ��

�

� �which, on using the orthogonality property, gives the non-vanishing coefficients as

01

,3

B � � 24

3B �

Substituting these values of A0 and A2 into Eq. (2.101), we obtain

21 2

1 4( , ) (cos )

3 3u r r Pθ θ� � �

which gives the potential everywhere inside the sphere. Similarly, substituting the values ofB0 and B2 into Eq. (2.102), we get



2 231 4( , ) (cos )3 3

u r Pr r

θ θ= − +

which gives the potential outside the sphere.

EXAMPLE 2.9 Find a general spherically symmetric solution of the following Helmholtzequation:

2 2( ) 0k u∇ − =

Solution In spherical polar coordinates, the Helmholtz equation can be written as

2 2 22

2 2 2 2 2 2 22 1 cot 1 0

sinu u u u u k u

r rr r r r∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂ ∂+ + + + − =θ

θθ θ φ(2.103)

In view of spherical symmetry, we look for u to be a function of r alone. Hence, Eq. (2.103)becomes

22

22 0u u k ur rr

∂ ∂∂∂

+ − =

Therefore, we have to solve

22 2 2

2 2 0u ur r k r urr

∂ ∂∂∂

+ − = (2.104)

Let

1 ( )u F rr

=

Differentiating twice with respect to r and rearranging, we obtain

22 1/2 1/2 3/2

2

( )2 2 ( )

3 ( ) ( ) ( )4

u F rr rF rr r

ur r F r r F r r F rr

∂∂

∂∂

−

= − + ′

= − − +′ ′′

Substituting the above relations, Eq. (2.104) becomes

2 2 2 1( ) ( ) ( ) 04

r F r rF r k r F r⎛ ⎞+ − + =′′ ′ ⎜ ⎟⎝ ⎠

or

22 2 2 1( ) ( ) ( ) ( ) 0

2r F r rF r ik r F r

⎡ ⎤⎛ ⎞+ + − =′′ ′ ⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦



This is the Bessel equation whose solution is

1/2 1/2( ) ( ) ( )F r AJ ikr BY ikr� �

where 1/2 1/2,J Y are Bessel functions with imaginary arguments, and is rewritten as

1/2 1/2( ) ( ) ( )F r AI kr BK kr� �

Therefore,

1/21/2 1/2( ) [ ( ) ( )]u r r AI kr BK kr��

But as ,r �� the solution should be finite, which is possible only if A = 0. It is also knownthat for large z,

1/2 ( )2

zK z ez

π ��

Thus the acceptable spherically symmetric solution of the Helmholtz equation is given by

1/2( )2

kr krcu r Br e e

kr r

π� � ��

where

2c B

k

π�

��

EXAMPLE 2.10 Show that the velocity potential for an irrotational flow of an incompressiblefluid satisfies the Laplace solution.

Solution Let us consider a closed surface S enclosing a fixed volume V in the regionoccupied by a moving fluid as shown in Fig. 2.5.

V

S

dS

q

n∧

Fig. 2.5 Conservation of mass.



Let ρ be the density of the fluid. If n̂ is a unit vector in the direction of the normal to thesurface element dS and q the velocity of the fluid at that point, then the inward normal velocity

is ˆ( ).n� �q Hence the mass of the fluid entering per unit time through the element dS is ˆ( ) .n dS� �q It

follows therefore that the mass of the fluid entering the surface S in unit time is

ˆ( )S

n dSρ� �� q

Also, the mass of the fluid within S is

V

dVρ��So the rate at which the mass goes on increasing is given by

V V

dV dVt t

��

� �

By conservation of mass, the rate of generation of mass within a given volume under theassumption that no internal sources are present is equal to the net inflow of mass through thesurface enclosing the given volume. Thus,

ˆ( )

div ( ) [using the divergence theorem]

V S

V

dV n dSt

dV

��

� � �

� �

��

��

q

q

ρ ρ

ρ

Therefore,

div ( ) 0V

dVt

��

� �� qρ ρ

Since the integrand is a continuous function and since this result is true for any arbitraryvolume element dV, it follows that the integrand is zero. Therefore,

0t

��

� � � �qρ ρ

which is called the equation of continuity. For an incompressible fluid, ρ � constant and,therefore,

0� � �q

Further, if the flow is irrotational, i.e., there exists a velocity potential φ such that

φ� ��qHence,

2 0φ φ� � � � �� q

Thus, an incompressible irrotational fluid satisfies the Laplace equation.



EXAMPLE 2.11 A thin rectangular homogeneous thermally conducting plate lies in the xy-

plane defined by 0 , 0 .x a y b� � � � The edge 0y � is held at the temperature ( ),Tx x a�where T is a constant, while the remaining edges are held at 0°. The other faces are insulatedand no internal sources and sinks are present. Find the steady state temperature inside theplate.

Solution Since no heat sources and sinks are present in the plate, the steady state

temperature u must satisfy 2 0.u� � Hence the problem is to solve

2PDE: 0u� �

BCs: (0, ) 0,u y � ( , ) 0,u a y � ( , ) 0,u x b � ( , 0) ( )u x Tx x a� �

This is a typical Dirichlet’s problem. The general solution satisfying the first three BCs isgiven by Eq. (2.47). Therefore,

1

( , ) sin sinh ( )nn

n nu x y A x y b

a a

π π�

�

� � � �� where

0

2sinh ( ) sin

a

nn b n

A f x x dxa a a

π π� � �� Using the last BC: ( , 0) ( ) ( ),u x Tx x a f x� � � we get

0

0

0

00

2 2

2sinh ( ) sin

2( ) sin

2( ) cos

2( ) cos (2 ) sin

2(2 ) sin

a

n

a

a

aa

n b nA Tx x a x dx

a a a

T nx x a x dx

a a

a T nx x a d x

n a a

T n a nx a x x a d x

n a n a

aT nx a x

an

� � ��

� ��

��

� ��

��

�

�

�

�

π π

π

ππ

π ππ π

ππ 0

0

2 20

2

2 2 3 3

2 sin

2 2sin cos

2 2 4(cos 1) [( 1) 1]

aa

a

n

nx dx

a

aT a na n x

n an

aT a a Tn

nn n

��

� ��

� � � � �

� π

ππππ

πππ π



Thus the required temperature distribution is given by

2

3 31

4( , ) cosech [( 1) 1] sin sinh ( )n

n

n Ta n nu x y b x y b

a a an

π π ππ

�

�

� � � � � �� EXAMPLE 2.12 Solve

2 0,u� � 0 ,x a� � 0 y b� �satisfying the BCs:

(0, ) 0,u y � ( , 0) 0,u x � ( , ) 0u x b �

3( , ) sinu y

a y Tx a

π��

Solution Using the variables separable method, one of the acceptable general solutionsis given by Eq. (2.38). Hence

1 2 3 4( , ) ( ) ( cos sin )px pxu x y c e c e c py c py��

Using the BC: ( , 0) 0,u x � we get

3 1 20 ( )px pxc c e c e��

implying c3 = 0. Therefore,

4 1 2( , ) sin ( )px pxu x y c py c e c e��

Now, using the BC: ( , ) 0,u x b � we obtain

4 1 20 sin ( )px pxc pb c e c e��

4 0c � (why?) implying sin 0pb � which gives

pb nπ� or ,n

pb

� π1, 2, 3,n � �

Thus,

4 1 2( , ) sin ( )px pxnu x y c y c e c e

b

π ��

Renaming the constants, we have

( , ) sin exp exp ,n n n

u x y y A x B xb b b

π π π� �� 1, 2,n � �

If we use the BC: (0, ) 0,u y � we get

0 sin ( )n

y A Bb

π� ��



giving A + B = 0; therefore, A = –B. Thus,

( , ) sin exp exp

2 sin sinh , 1, 2,

n n nu x y A y x x

b b b

n nA y x n

b b

� ��

� � � ��

π π π

π π

Differentiating with respect to x, we obtain

2 sin coshu n n n

A y xx b b b

π π π� � � ��

The last BC yields

3sin 2 sin coshy n n n

T A y aa b b b

π π π π� � � ��

from which we can determine 2A. Hence, the required solution is

3( , ) sin sech sinhbT y n n

u x y a xn a b b

� �� π π π

π

The principle of superposition gives the required solution as

3

1

( , ) sin sech sinhn

bT y n nu x y a x

n a b b

π π ππ

�

�

� � � ��

EXAMPLE 2.13 Find the potential function ( , , )u x y z in a rectangular box defined

by 0 ,x a� � 0 , 0y b z c� � � � (see Fig. 2.6), if the potential is zero on all sides and the

bottom, while ( , )u f x y� on the top of the box.

z

y

x

Ob

a

c

Fig. 2.6 Rectangular box.



Solution The potential distribution in the rectangular box satisfies the Laplace equation.Thus the problem is to solve

2 0xx yy zzu u u u� � � � �

subject to the BCs:

(0, , ) ( , , ) 0

( , 0, ) ( , , ) 0

( , , 0) 0

( , , ) ( , )

u y z u a y z

u x z u x b z

u x y

u x y c f x y

� �

� �

�

�

Following the variables separable method, let us assume the solution in the form

( , , ) ( ) ( ) ( )u x y z X x Y y Z z�

Substituting into the Laplace equation, we get

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0X x Y y Z z X x Y y Z z X x Y y Z z� � ��

which can also be written as

21

( ) ( ) ( )

( ) ( ) ( )

Y y Z z X x

Y y Z z X x

�� λ

where 1λ is a separation constant. Thus we have

21( ) ( ) 0X x X x� �� λ (2.105)

After the second separation, we also have

2 21 2

( ) ( )

( ) ( )

Z z Y y

Z z Y y

�� λ λ

22( ) ( ) 0Y y Y y� �� λ ��

23( ) ( ) 0Z z Z z� �� λ ��

where 2 2 23 1 2.λ λ λ� � The general solutions of Eqs. (2.105)–(2.107) are

1 1 2 1

3 2 4 2

5 3 6 3

( ) cos sin

( ) cos sin

( ) cosh sinh

X x c x c x

Y y c y c y

Z z c z c z

λ λ

λ λ

λ λ

� �

� �

� �



From the BCs,

1

1

(0) ( ) 0

(0) ( ) 0

(0) 0

(0) 0 gives 0

( ) 0 gives .

X X a

Y Y b

Z

X c

X a a mλ π

� �

� �

�

� �

� �

Therefore,

1,m

a� πλ 1, 2,m � �

Similarly,

3

2

(0) 0 gives 0

( ) 0 gives

Y c

Y b b nλ π

� �

� �

Therefore,

2,n

b� πλ 1, 2,n � �

Also, (0) 0Z � gives 5 0.c � Further, we note that

2 22 2 2 2 23 1 2 2 2

(say)mnm n

a bλ λ λ π λ

� ��

Then

2 2

3 2 2 mnm n

a bλ π λ� � �

The solutions now take the form

2

4

6

,( ) sin 1, 2,

,( ) sin 1, 2,

( ) sinh

m

n

mn mn

m xX x c m

a

n yY y c n

b

Z z c z

� � �

� � �

�

π

π

λ

Let 2 4 6 ;mn m n mnc c c c� then, after using the principle of superposition, the required solution is

1 1

( , , ) ( ) ( ) ( ) sin sin sinhmn mnm n

m x n yu x y z X x Y y Z z c z

a b

π π λ� �

� ��



Using the final BC: ( , ) ( , , ),f x y u x y c= we get

( , ) sinh sin sinmn mnm x n yf x y c c

a bπ πλ=∑ ∑

which is a double Fourier sine series. Thus, we have

0 0

4sinh ( , ) sin sina b

mn mnm x n yc c f x y dx dy

ab a bπ πλ = ∫ ∫ (2.109)

Therefore, Eqs. (2.108) and (2.109) constitute the required potential.

EXAMPLE 2.14 Find the electrostatic potential u in the annular region bounded by theconcentric spheres r = a, r = b, 0 < a < b (see Fig. 2.7), if the inner and outer surfaces are kept

at constant potentials u1 and u2, 1 2.u u≠

Solution The electrostatic potential satisfies the Laplace equation

2 0u∇ =

It is natural that we choose spherical polar coordinates. From the problem, it is evident thatwe are looking for a solution with spherical symmetry which is independent of θ and .φHence, ( ).u u r=

b

au u = 1

u u = 2

Fig. 2.7 Annular region.

Thus, we have to solve

2PDE: 0urr r∂ ∂∂ ∂

⎛ ⎞ =⎜ ⎟⎝ ⎠(2.110)

subject to

1

2

BCs: at

at

u u r a

u u r b

= =

= =

Integrating Eq. (2.110) with respect to r, we obtain

2 ur Ar

∂∂

= (a constant of integration)



Again, integrating, we get

Au B

r� � �

Now, using the BCs, we have

1 ,A

u Ba

� � � 2A

u Bb

� � �

Solving these equations, we get

2 1 ,(1/ ) (1/ )

u uA

a b

�

1 2( / ) ( / )

(1/ ) (1/ )

u b u aB

b a

�

Hence, using these values, the required potential is

1 21 1 1 1 1

(1/ ) (1/ )u u u

a b r b a r

� ��

EXAMPLE 2.15 A thermally conducting solid bounded by two concentric spheres of radii

a and b as shown in Fig. 2.8, a < b, is such that the internal boundary is kept at 1( )f θ and

the outer boundary at 2 ( ).f θ Find the steady state temperature in the solid.

Solution It is known that the steady temperature T satisfies the Laplace equation. Inthe present problem,

( , )T T r θ�

b

a r

O

P

�

x

y

z

Fig. 2.8 Region bounded by two concentric spheres.

Thus, we have to solve

2PDE: 0T� �



subject to the boundary conditions

1

2

( ) at

( ) at

T f r a

T f r b

θ

θ

� �

� �

In spherical polar coordinates, for axially symmetric case, the solution of the Laplace equationis given by Eq. (2.99) as follows:

10

( , ) (cos )n nn nn

n

BT r A r P

rθ θ

�

��

� �� Using the BCs:

1 10

( ) (cos )n nn nn

n

Bf A a P

aθ θ

�

��

� �� (2.111)

2 10

( ) (cos )n nn nn

n

Bf A b P

bθ θ

�

��

� �� (2.112)

In order to find the coefficients An and Bn, we have to express 1( )f θ and 2 ( )f θ in terms

of Legendre polynomials and compare the coefficients. In this process, the following orthogonalityrelation is useful:

0

0, if

(cos ) (cos ) sin 2 , if2 1

m n

m n

P P dm n

n

��

��

π

θ θ θ θ

Thus, multiplying both sides of Eq. (2.111) by (cos ) sinmP θ θ and integrating, we obtain

1 10 00

1

( ) (cos )sin (cos ) (cos )sin

2

2 1

n nm n n mn

n

m mm m

Bf P d A a P P d

a

BA a

ma

π πθ θ θ θ θ θ θ θ

�

��

�

� ��

� ��

��

(2.113)

Similarly, Eq. (2.112) gives

2 10 00

1

( ) (cos )sin (cos ) (cos )sin

2

2 1

n nm n n mn

n

m mm m

Bf P d A b P P d

b

BA b

mb

π πθ θ θ θ θ θ θ θ

�

��

�

� ��

� ��

��

��



Let

10

20

2 1 ( ) (cos ) sin2

2 1 ( ) (cos ) sin2

m m

m m

m f P d C

m f P d D

π

π

θ θ θ θ

θ θ θ θ

+ =

+ =

∫

∫Then Eqs. (2.113) and (2.114) reduce to

1

1

m mm mm

n mm mm

BA a Ca

BA b Db

+

+

+ =

+ =

Solving this pair of equations, we obtain

1 1

2 1 2 1

m mm m

m m mC a D b

Aa b

+ +

+ +−

=−

(2.115)

1 1

2 1 2 1( )m m m m

m mm m m

a b C b D aB

b a

+ +

+ +−

=−

(2.116)

Hence, the required steady temperature is

10

( , ) (cos )m mm mm

m

BT r A r P

rθ θ

∞

+=

⎛ ⎞= +⎜ ⎟⎝ ⎠∑

where Am and Bm are given by Eqs. (2.115) and (2.116).

EXAMPLE 2.16 A thin annulus occupies the region 0 , 0 2 .a r b< ≤ ≤ ≤ ≤θ π The faces are

insulated. Along the inner edge the temperature is maintained at 0°, while along the outer

edge the temperature is held at cos ( /2),T K θ= where K is a constant. Determine the temperature

distribution in the annulus.

Solution Mathematically, the problem is to solve2PDE: 0, , 0 2

BCs: ( , ) 0

( , ) cos /2

T a r b

T a

T b k

∇ = ≤ ≤ ≤ ≤

=

=

θ π

θ

θ θ

The required general solution is given by Eq. (2.57) in the form

1 2 3 4( , ) ( ) ( cos sin )n nT r c r c r c n c nθ θ θ−= + +



Using the first BC, we get

1 2 3 40 ( ) ( cos sin )n nc a c a c n c nθ θ��

implying thereby 1 2 0,n nc a c a�� or 22 1 .nc c a� � After adjusting the constants suitably, we have

2

( , ) ( cos sin )n

nn

aT r r A n B n

rθ θ θ

� ��

The principle of superposition gives

2

1

( , ) ( cos sin )n

nn nn

n

aT r r A n B n

rθ θ θ

�

�

� ��

��Now, using the second boundary condition, we obtain

2

1

( , ) cos ( ) ( cos sin )2

n n nn n

n

T b K b b a A n B nθθ θ θ

��

��

which is a full-range Fourier series. Hence,22

0

2

0

2

0

1( ) cos cos

2

1 1cos cos

2 2 2

1 1sin sin

2 21 122 2

0

n n nnA b b a K n d

kn n d

n nk

n n

π

π

π

θ θ θπ

θ θ θπ

θ θ

π

��

� ��

� ��

�

�

�

implying 0.nA � Also,

22

0

2

0

2

0

( ) cos sin2

1 1sin sin

2 2 2

1 1cos cos

2 21 122 2

n n nn

kB b b a n d

kn n d

n nk

n n

��

� ��

� ��

�

�

π

π

π

θ θ θπ

θ θ θπ

θ θ

π



2

1 1 1 11 1 1 122 2 2 2

1 1 21 1 12 2 4

k

n n n n

k k n

n n n

� ��

� � � ��

� ��

� � ��

π

π π

or

22

8( )

(4 1)n n n

nkn

B b b anπ

��

Thus the temperature distribution in the annulus is given by

21

8 ( / ) ( / )( , ) sin

4 1 ( / ) ( / )

n n

n nn

k n r a a rT r n

n b a a bθ θ

π

�

��

��

� ��

EXAMPLE 2.17 V is a function of r and θ satisfying the equation

2 2

2 2 2

1 10

V V V

r rr r

� � ��

� � �θ

within the region of the plane bounded by , , 0, /2.r a r b θ θ π� � � � Its value along the

boundary r a� is ( /2 ),θ π θ� along the other boundaries is zero. Prove that

4 2 4 2

4 2 4 2 31

2 ( / ) ( / ) sin (4 2)

( / ) ( / ) (2 1)

n n

n nn

r b b r nV

a b b a n

θπ

� � �

� ��

��

� ��

Solution The task is to solve the PDE

2 2

2 2 2

1 10

V V V

r rr r

� � ��

� � �θ

subject to the following boundary conditions:

(i) ( , ) 0,V b θ � 0 / 2θ π� �

(ii) ( , / 2) 0,V r π � a r b� �

(iii) ( , 0) 0,V r � a r b� �

(iv) ( , ) ( / 2 ),V a θ θ π θ� � 0 /2.θ π� �



The three possible solutions (see Section 2.8) are given as follows:

1 2 3 4

1 2 3 4

1 2 3 4

( ) ( cos sin )

[ cos ( ln ) sin ( ln )] ( )

( ln ) ( )

p p

p p

V c r c r c p c p

V c p r c p r c e c e

V c r c c c

θ θ

θ θ

θ

�

�

� � �

� � �

� � �

Since the problem is not defined for r = 0, ,� the second and third solutions are not acceptable.Hence, the generally acceptable solution is the first one. The boundary condition (iii) gives

3 1 20 ( )p pc c r c r��

implying 3 0.c � The boundary condition (ii) implies

4 1 20 sin ( )2

p pc p c r c rπ ��

Therefore,

sin 02

pπ � or 2 ,p n� 1, 2,n � �

Thus, the possible solution of the given equation has the form

2 24 1 2( , ) sin (2 ) ( )n nV r c n c r c rθ θ ��

Now, applying the boundary condition (i), we get

2 24 1 20 sin (2 ) ( )n nc n c b c bθ ��

which gives 42 1 .nc c b� � Therefore,

2 2 41 4( , ) sin (2 )[ ]n n nV r c c n r r bθ θ ��

Superposing all the solutions, we obtain

2 2 4

1

( , ) sin (2 ) ( )n n nn

n

V r c n r r bθ θ�

�

��

Satisfying boundary conditions (iv), we get

4 4

2sin (2 )

2

n n

n n

a bc n

a

πθ θ θ� ��

which is a Fourier sine series. Thus, we have

4 4/2

20

2sin (2 )

/2 2

n n

n n

a bn c

a

� �� π πθ θ θ

π



Integrating by parts, we obtain

/ 2 4 42

2 3 20

cos (2 ) sin (2 ) cos (2 )2 ( 2)

2 2 2 44 8

n n

n n

n n n a bc

n n n a

� ��

ππ θ π θ θ πθ θ θ

On simplification, we get

4 4

3 2

1{( 1) 1}

44

n nn

n n

a bc

n a

π � ��

Thus,

4 43

2

1 , for odd2

40, for even

n n

n n

na b nca

n

��

π

Hence, the required solution is

4 2 8 4 8 4

3 8 4 8 41

2 1( , ) sin (4 2)

(2 1)

n n n

n n

a r bV r n

rn a bθ θ

π

��

� �

� �� which can be recast in the form given in the problem.

EXAMPLE 2.18 Determine the potential of a grounded conducting sphere in a uniformfield defined by

2

0

PDE: 0, 0 , 0 , 0 2

BCs: (i) ( , ) 0.

(ii) cos as .

u r a

u a

u E r r

� � � � � � � �

�

� � � �

θ π φ π

θ

θ

Solution In spherical polar coordinates, with axial symmetry, the solution of the Laplaceequation is given by Eq. (2.99) in the form

10

( , ) (cos )n nn nn

n

Bu r A r P

rθ θ

�

��

� �� Using the boundary condition (ii), we have

00

( , ) (cos ) cosnn n

n

u r A r P E rθ θ θ�

��

which is true only for n = 1, when 1(cos ) cos .P θ θ� Also, 0nA � for 2.n � Therefore,

1 0( , ) cos cosu r A r E rθ θ θ� � �



implying 1 0.A E= − Hence,

0 11

( , ) cos (cos )nnn

n

Bu r E r P

rθ θ θ

∞

+=

= − +∑Now, applying the boundary condition (i), we get

0 11

0 cos (cos )nnn

n

BE a P

aθ θ

∞

+=

= − +∑

Multiplying both sides by (cos ) sinmP θ θ and integrating between the limits 0 to ,π we have

0 10 01cos ( ) (cos )sin (cos ) (cos ) sinn

m n mnn

BE a P d P P d

a

∞

+=

= ∑∫ ∫π π

θ θ θ θ θ θ θ θ (2.117)

Using the orthogonality property

0

0, for(cos ) (cos )sin 2 , for

2 1n m

m nP P d

m nm

≠⎧⎪= ⎨

=⎪⎩ +

∫π

θ θ θ θ

we obtain

01 0

2 cos ( ) (cos ) sin2 1

mmm

BE a P d

ma

πθ θ θ θ+ =

+ ∫or

20 0

2 1 cos ( ) (cos ) sin2

mm m

mB E a P dπ

θ θ θ θ++= ∫It can be verified that the integral on the right-hand side of the Eq. (2.117) vanishes forall m except when m = 1, in which case

31 0B E a=

Therefore, the required potential is given by3

00 2( , ) cos cos

E au r E r

rθ θ θ= − +

EXAMPLE 2.19 The steady, two-dimensional, incompressible viscous fluid flow past a circularcylinder, when the inertial terms are neglected (Stokes flow), is governed by the biharmonic

PDE: 4 0,ψ∇ = where ψ is the stream function. Find its solution subject to the BCs:

(i) ( , ) / 0r r∂ ∂= =ψ θ ψ on r = 1

(ii) ( , ) sinr rψ θ θ→ as .r →∞



Solution In view of the cylindrical geometry, we can write

4 2 2( ) 0ψ ψ� � � � �

where

2 22

2 2 2

1 1

r rr r

� � ��

� ��

ψ ψθ

Using the variables separable method, let us look for a solution of the form

( , ) ( ) sinr f rψ θ θ�

Therefore,

2 2

2 2

( ) sin , ( ) cos

( ) sin , ( ) sin

f r f rr

f r f rr

� ��

� ��

� ��

� � ��

ψ ψθ θθ

ψ ψθ θθ

Hence,

22

1 1( ) ( ) ( ) sinf r f r f r

r r

� �� ψ θ

which can also be written in the form

2 ( ) sinF rψ θ� �

where

2

1 1( ) ( ) ( ) ( )F r f r f r f r

r r� � ��

Therefore,4 2 2 2( ) [ ( ) sin ] 0F rψ ψ θ� � � � � � �

i.e.,

2

1 1( ) ( ) ( ) sin 0F r F r F r

r r

� �� θ

implying

2

1 1( ) ( ) ( ) 0F r F r F r

r r� � ��

Introducing the transformation , / ,zr e D d dz� � the above equation becomes

[ ( 1) 1] ( ) 0D D D F r� � � �



or2( 1) ( ) 0D F r� �

Its complementary function is

( ) z z BF r Ae Be Ar

r��

or

2

1 1( ) ( ) ( )

Bf r f r f r Ar

r rr� � � ��

i.e.2 3( ) ( ) ( )r f r rf r f r Ar Br� � � ��

which is a homogeneous ordinary differential equation. Again using the transformation ,zr e�

/ ,D d dz� we get

3[ ( 1) 1] z zD D D f Ae Be� � � � �or

2 3( 1) z zD f Ae Be� � �Its complementary function is

( ) z zf r Ce De�� while its particular integral is

33

2

1( )

8 21

z zz z Ae Bze

Ae BeD

� � ��

Therefore,

3( ) ln8 2

D A Bf r Cr r r r

r� � � �

Thus, we have

3 ln sin8 2

A B Dr r r Cr

rψ θ� ��

Now to satisfy the BC: sin , asr rψ θ� � � and from physical considerations, we choose

A = 0, Therefore,

ln sin2

B Dr r Cr

rψ θ ��



The boundary condition 0 on 1rψ � � gives ( ) sin 0,C D θ� � implying .C D� � Also, the

boundary condition / 0r� � �ψ on r = 1 gives

0 22 2

B BC D D� � � � �

implying B = 4D. Hence, the general solution is

12 ln sin

2 2

rD r r

rψ θ ��

EXAMPLE 2.20 The problem of axisymmetric fluid flow in a semi-infinite or in a finitecircular pipe of radius a is described as follows in cylindrical coordinates:

2PDE: 0, 0

BCs: (i) 0 at 0

(ii) 0, ( ) at

u r a

ur

r

u uV z r a

z r

� � � �

� �

� � �

��

� ��

Show that the speed of suction is given by

11

( ) ( cosh sinh ) ( )n n n n n nn

V z A z B z J aα α α α�

��

Solution In cylindrical coordinates ( , , ),r zθ

2 22

2 2 2 2

1 10

u u u uu

r rr r z

� � � ��

��

θIn axisymmetric case, the above equation becomes

2 2

2 2

10

u u u

r rr z

� � ��

� � � (2.118)

Let ( , ) ( ) ( )u r z f r zφ� which, when substituted into Eq. (2.118), gives

2(1/ )(say)

f r f

f

�� φ αφ

Then2 0� ��φ α φ ��

210f f f

r� � �� α (2.120)



The solution of Eq. (2.119) is

cosh sinhA z B zφ α α� �

Equation (2.120) can be rewritten as

2 2 2 0r f rf r f� � �� α

which is a Bessel’s equation of zeroth order whose general solution is

0 0( ) ( )f J r DY rα α� �

Here, 0 ( )J rα and 0 ( )Y rα are zeroth order Bessel functions of first and second kind respectively.

Therefore, the typical solution is

0 0( cosh sinh ) [ ( ) ( )]u A z B z J r DY rα α α α� � �

Now, 0 ( )Y rα is infinite at r = 0, and hence D = 0. Therefore, the possible solution is

0( cosh sinh ) ( )u A z B z J rα α α� �

The condition / 0u r� � � at r = 0 is automatically satisfied, since 0 ( ) 0J rα �� at r = 0. Now the

boundary condition / 0u z �� at r = a gives 0 ( ) 0,J aα � implying that aα are the zeros of the

Bessel function J0. Let these zeros be ( 0,1, 2, ).na n � �α Thus the appropriate solution is

01

( , ) ( cosh sinh ) ( )n n n n n nn

u r z A z B z J rα α α α�

��

Using the fact that, 0 1,J J� �� the speed of suction is given by

11

( ) ( cosh sinh ) ( )n n n n n nr a n

uV z A z B z J a

r

��

�

� �

� �� α α α α

EXAMPLE 2.21 Solve the following Poisson equation:

2 2

2 22

u u

x y

� ��

� �

subject to the boundary conditions

(0, ) (5, ) ( , 0) ( , 4) 0u y u y u x u x� � � �

Solution We assume the solution of the form

u v �� (2.121)

where v is a particular solution of the Poisson equation and ω is the solution of the correspondinghomogeneous Laplace equation. That is,



2 2v∇ = (2.122)

2 0ω∇ = (2.123)

It is customary to assume that v has the form

2 2( , )v x y a bx cy dx exy fy= + + + + +

Substituting this into Eq. (2.122), we get

2 2 2d f+ =

Let 0.f = Then 1.d = The remaining coefficients can be chosen arbitrarily. Thus we take

2( , ) 5v x y x x= − + (2.124)

so that v reduces to zero (satisfies the boundary conditions) on the sides x = 0 and x = 5.Now, we shall find ω from

2 0,ω∇ = 0 5,x< < 0 4y< < (2.125)

satisfying

2

2

(0, ) (0, ) 0

(5, ) (5, ) 0

( , 0) ( , 0) ( 5 )

( , 4) ( , 4) ( 5 )

y v y

y v y

x v x x x

x v x x x

ω

ω

ω

ω

= − =

= − =

= − = − − +

= − = − − +

The above conditions are obtained by using Eqs. (2.121), (2.124) and the given boundaryconditions. By using the superposition principle (see Section 2.5), the general solution ofEq. (2.125) is found to be

1( , ) sin ( /5) [ exp ( /5) exp ( /5)]n n

nx y n x a n y b n yω π π π

∞

== + −∑ (2.126)

Now, applying the non-homogeneous BC: 2( , 0) ( 5 ),x x xω = − − + we get, after renaming the

constants, the equation

2( , 0) ( 5 ) sin ( /5)nx x x A n xω π= − − + =∑Also, applying the BC: 2( , 4) ( 5 ),x x xω = − − + Eq. (2.126) can be rewritten in the form

2

1

4 4( 5 ) cosh sinh sin5 5 5n n

n

n n n xx x a bπ π π∞

=

⎛ ⎞− − + = +⎜ ⎟⎝ ⎠∑ (2.127)



which gives5 2

0

2(5 ) sin

5 5nn x

a x x dxπ� ��

Now, integrating by parts, the right-hand side yields

52 3

22 2 3 3

0

3 3

3 3 3 3

2 2

3 3 3 3 3 3

2 5 5 5(5 ) cos (5 2 ) sin ( 2) cos

5 5 5 5

2 2(5 ) 2 5cos

5

4(5 ) 1 cos 4(5) 1 ( 1)

n

n

n n x na x x x x x

n n n

nn n

n

n n n n

π π ππ π π

ππ π

ππ π

� ��

� ��

� ��

Hence,

2

3 3

8(5 ) , when is odd

0, when is even

n

na n

n

��

� ��

π (2.128)

Also, from Eq. (2.127), we have

5 2

0

4 4 2cosh sinh (5 )sin

5 5 5 5n n nn n n

a b x x x dx aπ π π ��

Therefore,

41 cosh

54

sinh5

n

n

a n

bn

π

π

� ��

(2.129)

Substituting an, bn from Eqs. (2.128) and (2.129) into Eq. (2.126), we get

1

4 4( , ) cosh sinh cosh sinh

5 5 5 5

4sinh sin sinh

5 5 5

nn

n nx y a y n n y

n ny x n

� �

� � �

� ��

�

�

� � � � � � � � ��

��

�



or

1

4( , ) sinh (4 ) sinh sin sinh5 55 5n

n n nx y y y x nπ π πω π∞

=

⎡ ⎤⎧ ⎫ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎨ ⎬ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎩ ⎭⎣ ⎦∑ (2.130)

Combining Eqs. (2.124), (2.128) and (2.130), the solution of the given Poisson equation isu(x, y) =

2

3 31

8 5 sinh (2 1) (4 )/5 sinh [(2 1) /5] sin [(2 1) /5]( 5)4 (2 1)sinh (2 1)5

n

n y n y n xx xnn

π π ππ π

∞

=

⎡ ⎤⎢ ⎥ ⎧ ⎫× − − + − −⎪ ⎪⎢ ⎥− + × ⎨ ⎬⎡ ⎤ −⎢ ⎥ ⎪ ⎪⎩ ⎭−⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

∑

EXAMPLE 2.22 Let IR be a region bounded by IR.∂ Let ( , , )P x y z be any point in the

interior of IR, as shown in Fig. 2.9. Let φ be a harmonic function in IR; also, let 1/ ,rψ = wherer is the distance from P. Applying Green’s second identity, show that

IR

1 1 1( )4

P dsr n n r

∂

∂ ∂∂ ∂

⎡ ⎤⎛ ⎞= − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∫∫ φφ φπ

y

x

z

O

IR

Pε

Σε

Q

∂Σε

∂IR


Solution Since ψ possesses a point of discontinuity in IR at ( , , ),P x y z Green’s second

identity cannot be directly applied to φ and .ψ However, 1/rψ = is bounded in IR εΣ− with

the boundary IR ,∂ ∂∪ εΣ where εΣ is a sphere of radius ε with centre at P. Now applying

Green’s second identity (2.19) to functions φ and ψ in IR– ,εΣ we get

2 2

IR IR

1(1/ ) (1/ ) (1/ )

(1/ ) (1/ )

r dV r r dSr n n

r dS r dSn n

∂

∂ ∂

∂ ∂∂ ∂

∂ ∂∂ ∂

−

⎡ ⎤ ⎡ ⎤∇ − ∇ = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎛ ⎞+ − ⎜ ⎟⎝ ⎠

∫∫∫ ∫∫

∫∫ ∫∫

ε

ε ε

φφ φ φ

φφ

Σ

Σ Σ

(2.131)



From the right-hand side of Eq. (2.131), we observe that the last two integrals depend only

on .ε But in the direction of the exterior normal to ,∂ εΣ we find that

21(1/ ) (1/ )

rr r

n r∂

∂ ∂∂ ∂ =

= − =ε ε εΣ

Therefore,

2

2 21 4(1/ ) ( ) 4 * ( )r dS dS Q Q

n∂ ∂

∂∂

= = =∫∫ ∫∫ε ε

πεφ φ φ πφε εΣ Σ

where * ( )Qφ is the average value of ( )Qφ on .εΣ∂ Further, the third integral

1 *(1/ ) 4r dS dSn n n

∂ ∂

∂ ∂ ∂∂ ∂ ∂

⎛ ⎞ ⎛ ⎞− = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫∫ ∫∫ε ε

φ φ φπεε

Σ Σ

where ( / ) *n∂ ∂φ is an average value of the normal derivative on .∂ εΣ Substituting these results

and using the fact that 2 (1/ ) 0r∇ = in IR ,εΣ− we obtain

2

IR IR

( 1/ ) (1/ ) (1/ ) 4 * ( ) 4 *r dV r r dS Pn n n

∂

∂ ∂ ∂∂ ∂ ∂

−

⎡ ⎤ ⎛ ⎞− ∇ = − + + ⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦∫∫∫ ∫∫ε

φ φφ φ πφ πεΣ

(2.132)

Now, taking the limit as 0,ε → and using the fact that φ is harmonic in IR ,εΣ− we arrive

at the fundamental result

IR

1( ) (1/ ) (1/ )4

P r r dSn n

∂

∂ ∂∂ ∂

⎡ ⎤= −⎢ ⎥⎣ ⎦∫∫ φφ φπ

(2.133)

Thus, the value of a harmonic function at any point of IR can be obtained in terms of the

values of φ and / n∂ ∂φ on the boundary IR∂ of the region IR .

EXAMPLE 2.23 Find the solution of the following Helmholtz equation, using separation ofvariables method:

—2u + K2u = uxx + uyy + uzz + K2u = 0 (2.134)

Solution It may be noted that the Laplacian in cartesian coordinates is a PDE with constantcoefficients, while in cylindrical or spherical coordinates, it is a PDE with variable coefficients.Thus, let us assume the solution of the given Helmholtz equation in the form

u(x, y, z) = X(x) Y(y) Z(z)

where X(x) is a function of x alone, Y(y) is a function of y alone, Z(z) is a function of z only.Substituting into the given Helmholtz equation, we get

X≤(x) Y(y) Z(z) + X(x) Y≤(y) Z(z) + X(x) Y(y) Z≤(z) + K2X(x) Y(y) Z(z) = 0,




2( ) ( ) ( )0

( ) ( ) ( )

X x Y y Z zK

X x Y y Z z

� � �� .

This equation is satisfied iff

2 2 21 2 3

( ) ( ) ( ), ,

( ) ( ) ( )

X x Y y Z zK K K

X x Y y Z z

� � �� (2.135)

and 2 2 2 21 2 3K K K K� � �

The sign of these separation constants K1, K2 and K3 need not be same, of course dependson the physical considerations. The solution of the three ODEs in Eq. (2.135) can be writtenin the form

1 1

2 2

3 3

1 2

3 4

5 6

( )

( )

( )

iK x iK x

iK y iK y

iK z iK z

X x C e C e

Y y C e C e

Z z C e C e

�

�

�

��

� � ��

(2.136)

Hence, in general, the solution of Eq. (2.134) can be written as

u(x, y, z) = AeiKr + Be–iKr.

However, if K2 is positive, the solution is of the form

u(x, y, z) = A cos(K1x + K2y + K3z) + B sin(K1x + K2y + K3z),

while, if K2 is negative, the solution is found to be

u(x, y, z) = A cosh(K1x + K2y + K3z) + B sinh(K1x + K2y + K3z).

��

1. Solve the following boundary value problem:

2

2

PDE: 0, 0 10, 0

400BCs: (10, ) ( )

( , 0) 0 ( , )

(0, ) is finite

u r

u

u r u r

u

� � � � � �

� �

� �

θ π

θ πθ θπ

π

θ

2. A homogeneous thermally conducting solid is bounded by the concentric spheresr = a, r = b, 0 < a < b. There are no heat sources within the solid. The inner surfacer = a is held at constant temperature T1, and at the outer surface there is radiationinto the medium r > b which is at a constant temperature T2. Find the steady temperatureT in the solid.



3. A thermally conducting solid bounded by two concentric spheres of radii a and b,a < b, is such that the internal boundary is kept at T1 and the outer boundaryat 2 (1 cos ).T θ� Find the steady state temperature in the solid.

4. A thin annulus occupies the region 0 , 0 2 ,a r b θ π� � � � � where b > a. The facesare insulated, and along the inner edge, the temperature is maintained at 0°, whilealong the outer edge, the temperature is held at 100°. Find the temperature distributionin the annulus.

5. A thermally conducting homogeneous disc with insulated faces occupies theregion 0 r a� � in the xy-plane. The temperature u on the rim ,r a� is

, 0

0, 2

Cu

θ α

α θ π

� ��

where α is a given angle 0 2 .α π� � Find the series expression for temperature atinterior points of the disc. In particular, consider the case when 100, /2.C α π� �

6. If ψ is a harmonic function which is zero on the cone θ α� and takes the value nnrαΣ

on the cone ,θ β� show that, when ,α θ β� �

0

(cos ) (cos ) (cos ) (cos )

(cos ) (cos ) (cos ) (cos )nn n n n

nn n n nn

Q P P Qr

Q P P Q

α θ α θψ αα β α β

�

�

��

7. Show that

,| |

q�� r r

ψ (q is constant)

is a solution of the Laplace equation.8. Solve the following

2 2

2 2 2

1 1PDE: 0

BCs: 0 at

cos at

1sin

r

r

r rr r

v r ar

v U r

v Ur�

� � �

�

�

�

��

�

�

�

� � �

� � �

� � �

� � �

� � ��

��

��



9. In the theory of elasticity, the stress function ,ψ in the problem of torsion of a beamsatisfies the Poisson equation

2 2

2 2 2,x y

∂ ∂∂ ∂

+ = −ψ ψ 0 1, 0 1x y≤ ≤ ≤ ≤

with the boundary conditions 0ψ = on sides 0, 1, 0x x y= = = and on 1.y = Findthe stress function .ψ

10. For an infinitely long conducting cylinder of radius a, with its axis coincident withits z-axis, the voltage ( , )u r θ obeys the Laplace equation

2 0,u∇ = 0 , 0 2r θ π≤ ≤ ∞ ≤ ≤

Find the voltage ( , )u r θ for r a≥ if Lt ( , ) 0,r

u r θ→∞

= subject to the condition

0 sin 3r a

uur a

∂∂ =

= θ

11. Hadamard’s example:(a) Consider the Cauchy problem for the Laplace equation

0xx yyu u+ = (E11.1)

subject to1( , 0) 0, ( , 0) sin ,yu x u x nxn

= = where n is a positive integer. Show that

its solution is

21( , ) sinh sin nu x y ny nxn

(E11.2)

(b) Show that for large n, the absolute value of the initial data in (a) can be madearbitrarily small, while the solution (E11.2) takes arbitrarily large values even at thepoints (x, y) with | y | as small as we want.

(c) Let f and g be analytic, and let u1 be the solution to the Cauchy problem describedby

0xx yyu u+ =

subject to

( , 0) ( ), ( , 0) ( )yu x f x u x g x= = (E11.3)

and let u2 be the solution of the Laplace equation (E11.1) subject to( , 0) ( ), ( , 0)yu x f x u x= ( ) (1/ ) sin .g x n nx= + Show that

2 1 21( , ) ( , ) sinh sinu x y u x y ny nxn

− = (E11.4)



(d) Conclude that the solution to the Cauchy problem for Laplace equation does notdepend continuously on the initial data. In other words, the initial value problem(Cauchy problem) for the Laplace equation is not well-posed. It may be noted thata problem involving a PDE is well-posed if the following three properties are satisfied:

(i) The solution to the problem exists.

(ii) The solution is unique.

(iii) The solution depends continuously on the data of the problem.

Fortunately, many a physical phenomena give rise to initial or boundary or IBVPswhich are well-posed.

12. Find the solution of the following PDE using separation of variables methoduxx – uy + u = 0.


182

��

Parabolic Differential Equations

��

The diffusion phenomena such as conduction of heat in solids and diffusion of vorticity inthe case of viscous fluid flow past a body are governed by a partial differential equation ofparabolic type. For example, the flow of heat in a conducting medium is governed by theparabolic equation

div ( ) ( , , )T

C K T H T tt

��

� � � rρ ��

where ρ is the density, C is the specific heat of the solid, T is the temperature at a point with

position vector r, K is the thermal conductivity, t is the time, and ( , , )H T tr is the amount of

heat generated per unit time in the element dV situated at a point (x, y, z) whose positionvector is r. This equation is known as diffusion equation or heat equation. We shall nowderive the heat equation from the basic concepts.

Let V be an arbitrary domain bounded by a closed surface S and let .V V S� � Let ( , ,T x y

, )z t be the temperature at a point (x, y, z) at time t. If the temperature is not constant, heat

flows from a region of high temperature to a region of low temperature and follows theFourier law which states that heat flux q (r, t) across the surface element dS with normal n̂ isproportional to the gradient of the temperature. Therefore,

( , ) ( , )t K T t� � �q r r (3.2)

where K is the thermal conductivity of the body. The negative sign indicates that the heat flux

vector points in the direction of decreasing temperature. Let n̂ be the outward unit normalvector and q be the heat flux at the surface element dS. Then the rate of heat flowing outthrough the elemental surface dS in unit time as shown in Fig. 3.1 is

ˆ( )dQ n dS� �q (3.3)


PARABOLIC DIFFERENTIAL EQUATIONS 183

dS

qn∧

Fig. 3.1 The heat flow across a surface.

Heat can be generated due to nuclear reactions or movement of mechanical parts as in inertialmeasurement unit (IMU), or due to chemical sources which may be a function of position,

temperature and time and may be denoted by ( , , ).H T tr We also define the specific heat of

a substance as the amount of heat needed to raise the temperature of a unit mass by a unittemperature. Then the amount of heat dQ needed to raise the temperature of the elemental

mass dm dVρ= to the value T is given by .dQ C T dV= ρ Therefore,

V

Q C T dV= ∫∫∫ ρ

V

dQ TC dVdt t

ρ= ∫∫∫ ∂∂

The energy balance equation for a small control volume V is: The rate of energy storage inV is equal to the sum of rate of heat entering V through its bounding surfaces and the rateof heat generation in V. Thus,

( , ) ˆ ( , , )V S V

T tC dV ndS H T t dVt

∂∂

= − ⋅ +∫∫∫ ∫∫ ∫∫∫r q rρ (3.4)

Using the divergence theorem, we get

( , ) div ( , ) ( , , ) 0V

TC t t H T t dVt

ρ r q r r⎡ ⎤+ − =⎢ ⎥⎣ ⎦∫∫∫ ∂∂

(3.5)

Since the volume is arbitrary, we have

( , ) div ( , ) ( , , )T tC t H T tt

∂∂

= − +r q r rρ (3.6)

Substituting Eq. (3.2) into Eq. (3.6), we obtain

( , ) [ ( , )] ( , , )T tC K T t H T tt

∂∂

= ∇ ⋅ ∇ +r r rρ (3.7)



If we define thermal diffusivity of the medium as

K

Cα

ρ�

then the differential equation of heat conduction with heat source is

21 ( , ) ( , , )( , )

T t H T tT t

t K

��

� � �r rr

α(3.8)

In the absence of heat sources, Eq. (3.8) reduces to

2( , )( , )

T tT t

t

��

� �rrα (3.9)

This is called Fourier heat conduction equation or diffusion equation. The fundamental problemof heat conduction is to obtain the solution of Eq. (3.8) subject to the initial and boundaryconditions which are called initial boundary value problems, hereafter referred to as IBVPs.

��

The heat conduction equation may have numerous solutions unless a set of initial and boundaryconditions are specified. The boundary conditions are mainly of three types, which we nowbriefly explain.

Boundary Condition I: The temperature is prescribed all over the boundary surface. That

is, the temperature T(r, t) is a function of both position and time. In other words, ( , )T G t� r which

is some prescribed function on the boundary. This type of boundary condition is called theDirichlet condition. Specification of boundary conditions depends on the problem underinvestigation. Sometimes the temperature on the boundary surface is a function of positiononly or is a function of time only or a constant. A special case includes T(r, t) = 0 on thesurface of the boundary, which is called a homogeneous boundary condition.

Boundary Condition II: The flux of heat, i.e., the normal derivative of the temperature / ,T n� �is prescribed on the surface of the boundary. It may be a function of both position and time,i.e.,

( , )T

f tn

��

� r

This is called the Neumann condition. Sometimes, the normal derivatives of temperature maybe a function of position only or a function of time only. A special case includes

0T

n

��

� on the boundary

This homogeneous boundary condition is also called insulated boundary condition whichstates that the heat flow is zero.



Boundary Condition III: A linear combination of the temperature and its normal derivativeis prescribed on the boundary, i.e.,

( , )TK hT G tn

∂∂

+ = r

where K and h are constants. This type of boundary condition is called Robin’s condition. Itmeans that the boundary surface dissipates heat by convection. Following Newton’s law ofcooling, which states that the rate at which heat is transferred from the body to the surroundingsis proportional to the difference in temperature between the body and the surroundings, wehave

( )aTK h T Tn

∂∂

− = −

As a special case, we may also have

0TK hTn

∂∂

+ =

which is a homogeneous boundary condition. This means that heat is convected bydissipation from the boundary surface into a surrounding maintained at zero temperature.

The other boundary conditions such as the heat transfer due to radiation obeying thefourth power temperature law and those associated with change of phase, like melting, ablation,etc. give rise to non-linear boundary conditions.

3.3 ELEMENTARY SOLUTIONS OF THE DIFFUSION EQUATION

Consider the one-dimensional diffusion equation

2

21 ,T T

tx∂ ∂

∂∂=

α, 0x t−∞ < < ∞ > (3.10)

The function

21( , ) exp [ ( ) /(4 )]4

= - -T x t x tt

ξ απα (3.11)

where ξ is an arbitrary real constant, is a solution of Eq. (3.10). It can be verified easily as

follows:

22

2

2

1 ( ) 1 exp [ ( ) /(4 )]4 24

1 2( ) exp [ ( ) /(4 )]4 4

T x x tt t tt

T x x tx t t

ξ ξ απα α

ξ ξ απα α

−= − − −

− − = − −

∂∂

∂∂



Therefore,

2 22

2 2 2

1 1 ( ) 1exp [ ( ) /(4 )]

4 2 4

T x Tx t

t t tx t

� ��

��

ξ ξ απα α αα

which shows that the function (3.11) is a solution of Eq. (3.10). The function (3.11), knownas Kernel, is the elementary solution or the fundamental solution of the heat equation for theinfinite interval. For t > 0, the Kernel T(x, t) is an analytic function of x and t and it can alsobe noted that T(x, t) is positive for every x. Therefore, the region of influence for the diffusion

equation includes the entire x-axis. It can be observed that as | | ,x � � the amount of heat

transported decreases exponentially.In order to have an idea about the nature of the solution to the heat equation, consider

a one-dimensional infinite region which is initially at temperature f (x). Thus the problem isdescribed by

2

2PDE: , , 0

IC: ( , 0) ( ), , 0

T Tx t

t x

T x f x x t

α� ��

� ��

� ��

(3.12)

(3.13)

Following the method of variables separable, we write

( , ) ( ) ( )T x t X x tβ� (3.14)

Substituting into Eq. (3.12), we arrive at

1X

X

�� β λα β

(3.15)

where λ is a separation constant. The separated solution for β gives

tCeαλβ � (3.16)

If 0,λ � we have β and, therefore, T growing exponentially with time. From realistic physical

considerations, it is reasonable to assume that ( ) 0f x � as | | ,x � � while | ( , )|T x t M� as

| | .x � � But, for T(x, t) to remain bounded, λ should be negative and thus we

take 2.λ μ� � Now from Eq. (3.15) we have

2 0X X� �� μIts solution is found to be

1 2cos sinX c x c xμ μ� �Hence

2( , , ) ( cos sin ) tT x t A x B x e αμμ μ μ �� (3.17)



is a solution of Eq. (3.12), where A and B are arbitrary constants. Since f (x) is in general notperiodic, it is natural to use Fourier integral instead of Fourier series in the present case. Also,since A and B are arbitrary, we may consider them as functions of μ and take

( ), ( ).A A B Bμ μ� � In this particular problem, since we do not have any boundary conditions

which limit our choice of ,μ we should consider all possible values. From the principles ofsuperposition, this summation of all the product solutions will give us the relation

2

0 0( , ) ( , , ) [ ( )cos ( )sin ] tT x t T x t d A x B x e d

� � �� αμμ μ μ μ μ μ μ (3.18)

which is the solution of Eq. (3.12). From the initial condition (3.13), we have

0( , 0) ( ) [ ( )cos ( ) sin ]T x f x A x B x d

�� μ μ μ μ μ (3.19)

In addition, if we recall the Fourier integral theorem, we have

0

1( ) ( ) cos ( )f t f x t x dx dω ω

π� �

��

� �� (3.20)

Thus, we may write

0

1( ) ( ) cos ( )f x f y x y dy d

� �

��

� �� μ μπ

0

0

1( ) (cos cos sin sin )

1cos ( )cos sin ( ) sin

f y x y x y dy d

x f y y dy x f y y dy d

� �

��

� � �

��

� ��

� ��

� �

� � �

μ μ μ μ μπ

μ μ μ μ μπ

(3.21)

Let

1( ) ( ) cos

1( ) ( ) sin

A f y y dy

B f y y dy

μ μπ

μ μπ

�

��

�

��

�

�

�

�Then Eq. (3.21) can be written in the form

0( ) [ ( ) cos ( )sin ]f x A x B x d

�� μ μ μ μ μ (3.22)

Comparing Eqs. (3.19) and (3.22), we shall write relation (3.19) as

0

1( , 0) ( ) ( ) cos ( )T x f x f y x y dy d

� �

��

� �� μ μπ

(3.23)



Thus, from Eq. (3.18), we obtain

2

0

1( , ) ( ) cos ( ) exp ( )T x t f y x y t dy d

� �

��

� �� μ αμ μπ

(3.24)

Assuming that the conditions for the formal interchange of orders of integration are satisfied,we get

2

0

1( , ) ( ) exp ( ) cos ( )T x t f y t x y d dyαμ μ μ

π� �

��

� �� (3.25)

Using the standard known integral

2 2

0exp ( ) cos (2 ) exp ( )

2s bs ds b

π�� (3.26)

Setting ,s t� μ α and choosing

2

x yb

tα��


2 2

0cos ( ) exp [ ( ) /(4 )]

4te x y d x y t

t

� � � � � �� αμ πμ μ αα

(3.27)

Substituting Eq. (3.27) into Eq. (3.25), we obtain

21( , ) ( ) exp [ ( ) /(4 )]

4T x t f y x y t dy

tα

απ�

�� (3.28)

Hence, if f (y) is bounded for all real values of y, Eq. (3.28) is the solution of the problemdescribed by Eqs. (3.12) and (3.13).

EXAMPLE 3.1 In a one-dimensional infinite solid, ,x�� the surface a x b� � isinitially maintained at temperature T0 and at zero temperature everywhere outside the surface.Show that

0( , ) erf erf2 4 4

T b x a xT x t

t tα α� � � ��

where erf is an error function.

Solution The problem is described as follows:

0

PDE: ,

IC: ,

0 outside the above region

t xxT T x

T T a x b

α� � � � � �

� � �

�



The general solution of PDE is found to be

21( , ) ( ) exp [ ( ) /(4 )]

4T x t f x t d

tξ ξ α ξ

πα�

��

Substituting the IC, we obtain

20( , ) exp [ ( ) /(4 )]4

b

a

TT x t x t d

tξ α ξ

πα� � ��

Introducing the new independent variable η defined by

4

x

t

ξηα��

and hence

4d t dξ α η�

the above equation becomes

2 2 2( ) / (4 ) ( ) / ) ( ) / )0 0

( ) / ) 0 0

2 2( , )

2

b x t b x t a x t

a x t

T TT x t e d e d e d

α α αη η ηα

η η ηπ π π

(4 (4√ √ √

(4√

� � ��

� ��

� �� Now we introduce the error function defined by

2

0

2erf ( ) exp ( )

zz dη η

π� ��

Therefore, the required solution is

0( , ) erf erf2 4 4

T b x a xT x t

t tα α� � � ��

��

According to the notion in mechanics, we come across a very large force (ideally infinite)acting for a short duration (ideally zero time) known as impulsive force. Thus we have afunction which is non-zero in a very short interval. The Dirac delta function may be thoughtof as a generalization of this concept. This Dirac delta function and its derivative play a usefulrole in the solution of initial boundary value problem (IBVP).

Consider the function having the following property:

1/2 , | |( )

0, | |

tt

tε

ε εδ

ε

��

(3.29)



Thus,

1( ) 1

2t dt dt

εε εδ

ε�

�� (3.30)

Let f (t) be any function which is integrable in the interval ( , ).ε ε� Then using the Mean-

value theorem of integral calculus, we have

1( ) ( ) ( ) ( ),

2f t t dt f t dt f

εε εδ ξ

ε�

�� ε ξ ε� � � (3.31)

Thus, we may regard ( )tδ as a limiting function approached by ( )tεδ as 0,ε � i.e.

0( ) Lt ( )t t

�� εε

δ δ (3.32)

As 0,ε � we have, from Eqs. (3.29) and (3.30), the relations

0

(in the sense of being very large)

, if 0( ) Lt ( )

0, if 0

tt t

t

εεδ δ

��

(3.33)

( ) 1t dtδ�

�� (3.34)

This limiting function ( )tδ defined by Eqs. (3.33) and (3.34) is known as Dirac delta function

or the unit impulse function. Its profile is depicted in Fig. 3.2. Dirac originally called it animproper function as there is no proper function with these properties. In fact, we can observethat

| |0 01 ( ) Lt ( ) Lt 0 0

tt dt t dtεεε ε

δ δ�

��

δε

1/2ε

–ε 0 εt

Fig. 3.2 Profile of Dirac delta function.

Obviously, this contradiction implies that ( )tδ cannot be a function in the ordinary sense.

Some important properties of Dirac delta function are presented now:



PROPERTY I: ( ) 1t dtδ�

��

PROPERTY II: For any continuous function f (t),

( ) ( ) (0)f t t dt fδ�

��

Proof Consider the equation

0 0Lt ( ) ( ) Lt ( ),f t t dt fεε ξ

δ ξ�

�� ε ξ ε� � �

As 0,ε � we have 0.ξ � Therefore,

( ) ( ) (0)f t t dt fδ�

��

PROPERTY III: Let f (t) be any continuous function. Then

( ) ( ) ( )t a f t dt f aδ∞

−∞− =∫

Proof Consider the function

1/ ,( )

0, elsewheree

a t at a

ε εδ

< < +⎧⎪− = ⎨⎪⎩

Using the mean-value theorem of integral calculus, we have

1( ) ( ) ( ) ( ),

a

at a f t dt f t dt f a

εεδ θε

ε� �

�� 0 1θ� �

Now, taking the limit as 0,ε � we obtain

( ) ( ) ( )t a f t dt f aδ�

��

Thus, the operation of multiplying f (t) by ( )t aδ � and integrating over all t is equivalent to

substituting a for t in the original function.

PROPERTY IV: ( ) ( )t tδ δ� �

PROPERTY V: 1( ) ( ),at t

aδ δ� 0a �



PROPERTY VI: If ( )tδ is a continuously differentiable. Dirac delta function vanishing for large

t, then

( ) ( ) (0)f t t dt f∞

−∞= −′ ′∫ δ

Proof Using the rule of integration by parts, we get

( ) ( ) [ ( ) ( )] ( ) ( )f t t dt f t t f t t dt∞ ∞∞

−∞−∞ −∞= −′ ′∫ ∫δ δ δ

Using Eq. (3.33) and property (III), the above equations becomes

( ) ( ) (0)f t t dt fδ∞

−∞= −′ ′∫

PROPERTY VII: ( ) ( ) ( )t a f t dt f aδ∞

−∞− = −′ ′∫

Having discussed the one-dimensional Dirac delta function, we can extend the definition totwo dimensions. Thus, for every f which is continuous over the region S containing the

point ( , ),ξ η we define ( , )x yδ ξ η− − in such a way that

( , ) ( , ) ( , )S

x y f x y d fδ ξ η σ ξ η− − =∫∫ (3.35)

Note that ( , )x yδ ξ η− − is a formal limit of a sequence of ordinary functions, i.e.,

0( , ) Lt ( )x y r

→− − = εε

δ ξ η δ (3.36)

where 2 2 2( ) ( ) .r x yξ η= − + − Also observe that

( ) ( ) ( , ) ( , )x y f x y dx dy fδ ξ δ η ξ η− − =∫∫ (3.37)

Now, comparing Eqs. (3.35) and (3.37), we see that

( , ) ( ) ( )x y x yδ ξ η δ ξ δ η− − = − − (3.38)

Thus, a two-dimensional Dirac delta function can be expressed as the product of two one-dimensional delta functions. Similarly, the definition can be extended to higher dimensions.

EXAMPLE 3.2 A one-dimensional infinite region x−∞ < < ∞ is initially kept at zero

temperature. A heat source of strength gs units, situated at x ξ= releases its heat instantaneously

at time .t τ= Determine the temperature in the region for .t τ>



Solution Initially, the region x�� is at zero temperature. Since the heat source

is situated at x ξ� and releases heat instantaneously at ,t τ� the released temperature

at x ξ� and t τ� is a -δ function type. Thus, the given problem is a boundary value problem

described by

2

2

( , ) 1PDE: , , 0

IC: ( , ) ( ) 0, , 0

( , ) ( ) ( )s

T g x t Tx t

k tx

T x t F x x t

g x t g x t

α

δ ξ δ τ

� � ��

� � ��

� � �

� ��

The general solution to this problem as given in Example 7.25, after using the initial condition

( ) 0,F x � is

2

0( , ) ( , ) exp [ ( ) /{4 ( )}]

4 ( )

t

t x

dtT x t g x t x x t t dx

k t t

α απα

�

� ��

��

Since the heat source term is of the Dirac delta function type, substituting

( , ) ( ) ( )sg x t g x tδ ξ δ τ� � �

into Eq. (3.39), and integrating we get, with the help of properties of delta function, therelation

2

0

exp [ ( ) /{4 ( )}]( , ) ( )

4

tsg x t t

T x t t dtk t t

α ξ α δ τπα

� � � ��

Therefore, the required temperature is

2exp [ ( ) /{4 ( )}]( , )

4 ( )sg x t

T x tk t

α ξ α τπα τ

� � ��

for t τ�

EXAMPLE 3.3 An infinite one-dimensional solid defined by x�� is maintained at

zero temperature initially. There is a heat source of strength ( )sg t units, situated at ,x ξ� which

releases constant heat continuously for t > 0. Find an expression for the temperature distributionin the solid for t > 0.

Solution This problem is similar to Example 3.2, except that ( , ) ( ) ( )sg x t g t xδ ξ� � is

a Dirac delta function type. The solution to this IBVP is

2

0

( )( , ) exp [ ( ) /{4 ( )}]

4 ( )

ts

t

g tT x t x t t dt

K t t��

��

� ��α ξ απα

(3.40)



It is given as ( ) constant (say).s sg t g� � Let us introduce a new variable η defined by

4 ( )

x

t t

ξηα�

��

or2

2

1( )

4

xt t

ξη α��

Therefore,

2

3

1 ( )

2

xdt d

ξ ηαη��

Thus, Eq. (3.40) becomes

2

2( )/ 4

exp ( )( , )

2sx t

xT x t g d

K ξ α

ξ η ηπ η

�

�

� ��

However,2 2

2

22

d e ee

d

η ηη

η η η

� ��

� ��

Hence,

22

( ) / (4 )( ) / 4

( , ) 22s

x tx t

x eT x t g e d

K

ηη

ξ αξ α

ξ ηπ η

��

��

� ��

Recalling the definitions of error function and its complement

2

0

2 2

0 0

2

2erf ( ) , erf ( ) 1

2erfc( ) 1 erf ( ) exp( ) exp( )

2exp( )

x

x

x

x e d

x x d d

d

η ηπ

η η η ηπ

η ηπ

�

�

�

� � �

� ��

� �

�

� �

�the temperature distribution can be expressed as

2 | |( , ) exp [ ( ) /(4 )] 1 erf

2 2 4sg t x x

T x t x tK t

α ξ ξξ απ α α

��

Alternatively, the required temperature is

2 | |( , ) exp [ ( ) /(4 )] erfc

2 2 4sg t x x

T x t x tK t

��

α ξ ξξ απ α α



��

Consider the equation

2

2

T T

t x

� ��

�α (3.41)

Among the many methods that are available for the solution of the above parabolic partialdifferential equation, the method of separation of variables is very effective and straightforward.We separate the space and time variables of T(x, t) as follows: Let

( , ) ( ) ( )T x t X x tβ� (3.42)

be a solution of the differential Eq. (3.41). Substituting Eq. (3.42) into (3.41), we obtain

1,

XK

X

�� βα β

a separation constant

Then we have

2

20

0

d XKX

dx

dK

dt

β α β

� �

� �

(3.43)

(3.44)

In solving Eqs. (3.43) and (3.44), three distinct cases arise:

Case I When K is positive, say 2,λ the solution of Eqs. (3.43) and (3.44) will have the form

1 2 ,x xX c e c eλ λ�� 2

3tc eαλβ � (3.45)

Case II When K is negative, say 2,λ� then the solution of Eqs. (3.43) and (3.44) will have

the form

1 2cos sin ,X c x c xλ λ� � 2

3tc e αλβ �� (3.46)

Case III When K is zero, the solution of Eqs. (3.43) and (3.44) can have the form

1 2 ,X c x c� � 3cβ � (3.47)

Thus, various possible solutions of the heat conduction equation (3.41) could be the following:

2

2

1 2

1 2

1 2

( , ) ( )

( , ) ( cos sin )

( , )

x x t

t

T x t c e c e e

T x t c x c x e

T x t c x c

λ λ αλ

αλλ λ

�

�

� ��

� ��

� ��

��



where

1 1 3,c c c�� 2 2 3c c c��

EXAMPLE 3.4 Solve the one-dimensional diffusion equation in the region 0 , 0,x tπ� � �subject to the conditions

(i) T remains finite as t ��(ii) 0, if 0 and for allT x tπ� �

(iii)

, 0 /2

At 0,, .

2

x x

t Tx x

πππ π

� ��

� � ��

Solution Since T should satisfy the diffusion equation, the three possible solutions are:

2

2

1 2

1 2

1 2

, ) (( )

( , ) ( cos sin )

( , ) ( )

x x t

t

t cT x e c e e

T x t c x c x e

T x t c x c

λ λ αλ

αλλ λ

�

�

� �

� �

� �

The first condition demands that T should remain finite as .t �� We therefore reject the firstsolution. In view of BC (ii), the third solution gives

1 20 0 ,c c� � � 1 20 c cπ� � �

implying thereby that both c1 and c2 are zero and hence T = 0 for all t. This is a trivialsolution. Since we are looking for a non-trivial solution, we reject the third solution also.Thus, the only possible solution satisfying the first condition is

2

1 2( , ) ( cos sin ) tT x t c x c x e αλλ λ �� Using the BC (ii), we have

1 2 00 ( cos sin )

xc x c xλ λ ��

implying c1 = 0. Therefore, the possible solution is2

2( , ) sintT x t c e xαλ λ��

Applying the BC: 0T � when ,x π� we get

sin 0 n� � �λπ λπ π

where n is an integer. Therefore,

nλ �Hence the solution is found to be of the form

2( , ) sinn tT x t ce nxα��



Noting that the heat conduction equation is linear, its most general solution is obtained byapplying the principle of superposition. Thus,

2

1

( , ) sinn tn

n

T x t c e nxα�

�

��

Using the third condition, we get

1

( , 0) sinnn

T x c nx�

��

which is a half-range Fourier-sine series and, therefore,

/2

0 0 /2

2 2( , 0) sin sin ( )sinnc T x nx dx x nx dx x nx dx

π π π

ππ

π π� ��


/2

2 20 /2

2 cos sin cos sin( )n

nx nx nx nxc x x

n nn n

π π

ππ

π

�� ! " �� or

2

4 sin ( /2)n

nc

n

ππ

�

Thus, the required solution is

2

21

4 sin ( /2)( , ) sin

n t

n

e nT x t nx

n

α ππ

� �

��

EXAMPLE 3.5 A uniform rod of length L whose surface is thermally insulated is initiallyat temperature � = �0. At time t = 0, one end is suddenly cooled to � = 0 and subsequentlymaintained at this temperature; the other end remains thermally insulated. Find the temperaturedistribution �(x, t).

Solution The initial boundary value problem IBVP of heat conduction is given by2

2

0

PDE: , 0 , 0

BCs: (0, ) 0, 0

( , ) 0, 0

IC: ( , 0) , 0

x L tt x

t t

L t tx

x x L

� ��

��

� � � �

� �

� �

� � �

θ θα

θ

θ

θ θ



From Section 3.5, it can be noted that the physically meaningful and non-trivial solution is

2( , ) ( cos sin )tx t e A x B x�� αλθ λ λ

Using the first boundary condition, we obtain 0.A � Thus the acceptable solution is

2

2

sin

cos

t

t

Be x

Be xx

��

�

�

�

�

αλ

αλ

θ λ

θ λ λ

Using the second boundary condition, we have2

0 costBe Lαλλ λ��

implying cos 0.Lλ � Therefore,The eigenvalues and the corresponding eigenfunctions are

(2 1),

2nn

L

πλ �� 0,1, 2,n � #

Thus, the acceptable solution is of the form

2 2 2 1exp [ {(2 1) / 2 } ] sin

2

nB n L t x

Lθ α π π��

Using the principle of superposition, we obtain

2 2

0

2 1( , ) exp [ {(2 1)/2 } ] sin

2nn

nx t B n L t x

Lθ α π π

�

�

�� Finally, using the initial condition, we have

00

2 1sin

2nn

nB x

Lθ π

�

�

�� which is a half-range Fourier-sine series and, thus,

00

00

0 0

2 2 1sin

2

2 2 2 1cos

(2 1) 2

4 4[cos {(2 1) /2} cos 0]

(2 1) (2 1)

L

n

L

nB x dx

L L

L nx

L n L

nn n

θ π

θ ππ

θ θππ π

��

��

� � � � ��

�



Thus, the required temperature distribution is

2 20

0

4 2 1( , ) exp [ {(2 1)/2 } ] sin

(2 1) 2n

nx t n L t x

n L

θθ α π ππ

�

�

��

EXAMPLE 3.6 A conducting bar of uniform cross-section lies along the x-axis with endsat 0x � and .x L� It is kept initially at temperature 0° and its lateral surface is insulated. Thereare no heat sources in the bar. The end x = 0 is kept at 0°, and heat is suddenly applied atthe end x = L, so that there is a constant flux q0 at x = L. Find the temperature distributionin the bar for t > 0.

Solution The given initial boundary value problem can be described as follows:

2

2

0

PDE:

BCs: (0, ) 0, 0

( , ) , 0

IC: ( , 0) 0, 0

T T

t x

T t t

TL t q t

x

T x x L

� ��

��

�

�

�

� � �

α

Prior to applying heat suddenly to the end x = L, when t = 0, the heat flow in the bar isindependent of time (steady state condition). Let

( ) 1( , ) ( ) ( , )sT x t T x T x t� �

where T(s) is a steady part and T1 is the transient part of the solution. Therefore,

2( )2

0sT

x

�

��

whose general solution is

( )sT Ax B� �

when ( )0, 0,sx T� � implying 0.B � Therefore,

( )sT Ax�

Using the other BC: ( )0 ,

sTq

x

��

� we get 0.A q� Hence, the steady state solution is

( ) 0sT q x�



For the transient part, the BCs and IC are redefined as

(i) 1 ( )(0, ) (0, ) (0) 0 0 0sT t T t T� � � � �

(ii) 1 0 0( , )/ ( , )/ ( , )/ 0sT L t x T L t x T L t x q q� � � � � ��

(iii) 1 ( ) 0( , 0) ( , 0) ( ) , 0 .sT x T x T x q x x L� � � � � �

Thus, for the transient part, we have to solve the given PDE subject to these conditions. Theacceptable solution is given by Eq. (3.48), i.e.

2

1( , ) ( cos sin )tT x t e A x B xαλ λ λ��

Applying the BC (i), we get 0.A � Therefore,

2

1( , ) sintT x t Be xαλ λ��

and using the BC (ii), we obtain

21 cos 0t

x L

TB e L

x

��

�

�� αλλ λ

implying (2 1) , 1, 2,2

L n n� � � �πλ Using the superposition principle, we have

2 21

1

2 1( , ) exp [ {(2 1)/2 } ]sin

2nn

nT x t B n L t x

Lα π π

�

�

��

Now, applying the IC (iii), we obtain

1 01

2 1( , 0) sin

2nn

nT x q x B x

Lπ

�

�

��

Multiplying both sides by sin2 1

2

mx

Lπ��

� �and integrating between 0 to L and noting that

0

0,2 1 2 1

sin sin2 2 ,

2

L

n m

n mn m

B x x dx B LL L n mπ π

��

��

we get at once, after integrating by parts, the equation

2

0 2 2

4 2 1sin

2 2(2 1)m

L m Lq B

mπ

π� � ��



or2

10 2 2

4( 1)

2(2 1)m

mL L

q Bm π

��

which gives

02 2

( 1) 8

(2 1)

m

mLq

Bm π�

��

Hence, the required temperature distribution is

2 200 2 2

1

8 ( 1) 2 1( , ) exp [ {(2 1)/ } ] sin

2(2 1)

m

m

Lq mT x t q x m L t x

Lmα π π

π

�

�

��

EXAMPLE 3.7 The ends A and B of a rod, 10 cm in length, are kept at temperatures 0°Cand 100°C until the steady state condition prevails. Suddenly the temperature at the end A isincreased to 20°C, and the end B is decreased to 60°C. Find the temperature distribution inthe rod at time t.

Solution The problem is described by2

2PDE: , 0 10

BCs: (0, ) 0, (10, ) 100

T Tx

t x

T t T t

� ��

� � �

� �

α

Prior to change in temperature at the ends of the rod, the heat flow in the rod is independentof time as steady state condition prevails. For steady state,

2

20

d T

dx�

whose solution is

( )sT Ax B� �

When 0, 0,x T� � implying 0.B � Therefore,

( )sT Ax�

When 10, 100,x T� � implying 10.A � Thus, the initial steady temperature distribution in

the rod is

( ) ( ) 10sT x x�

Similarly, when the temperature at the ends A and B are changed to 20 and 60, the final steadytemperature in the rod is

( ) ( ) 4 20sT x x� �



which will be attained after a long time. To get the temperature distribution ( , )T x t in the

intermediate period, counting time from the moment the end temperatures were changed, weassume that

1 ( )( , ) ( , ) ( )sT x t T x t T x= +

where 1( , )T x t is the transient temperature distribution which tends to zero as .t →∞ Now,

1( , )T x t satisfies the given PDE. Hence, its general solution is of the form

2( , ) (4 20) ( cos sin )tT x t x e B x c xαλ λ λ−= + + +

Using the BC: 20T = when x = 0, we obtain

220 20 tBe αλ−= +

implying B = 0. Using the BC: T = 60 when x = 10, we get

sin 10 0, implying ,10nπλ λ= = 1, 2,n = …

The principle of superposition yields

2

1( , ) (4 20) exp [ ( /10) ]sin

10nn

nT x t x c n t xπα π∞

=

⎛ ⎞= + + − ⎜ ⎟⎝ ⎠∑

Now using the IC: T = 10x, when t = 0, we obtain

10 4 20 sin10nnx x c xπ⎛ ⎞= + + ⎜ ⎟⎝ ⎠∑

or

6 20 sin10nnx c xπ⎛ ⎞− = ⎜ ⎟⎝ ⎠∑

where

10

0

2 1 800 200(6 20) sin ( 1)10 10 5

nn

nc x x dxn n

ππ π

⎛ ⎞ ⎡ ⎤= − = − − −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦∫Thus, the required solution is

2

1

1 800 200( , ) 4 20 ( 1) exp sin5 10 10

n

n

n nT x t x t xn n

π παπ π

∞

=

⎡ ⎤⎡ ⎤ ⎛ ⎞ ⎛ ⎞= + − − − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎢ ⎥⎣ ⎦∑ .



EXAMPLE 3.8 Assuming the surface of the earth to be flat, which is initially at zerotemperature and for times t > 0, the boundary surface is being subjected to a periodic heatflux g0 cos .tω Investigate the penetration of these temperature variations into the earth’ssurface and show that at a depth x, the temperature fluctuates and the amplitude of the steadytemperature is given by

0 2exp [ ( /2 ) ]

2

gx

α ω αω

�

Solution The given IBVP is described by

2

2

0

PDE: (3.49)

BC: cos at 0, 0 (3.50)

IC: ( , 0) 0 (3.51)

T T

t x

Tg t x t

x

T x

� ��

��

�

� � �

�

α

ω

We shall introduce an auxiliary function T� satisfying Eqs. (3.49)–(3.51) and then define thecomplex function Z such that

Z T iT� � �

We can easily verify that Z satisfies

2

2

0

PDE: (3.52)

BC: at 0, 0

IC: 0 in the region, 0

i t

Z Z

t x

Zg e x t

x

Z t

� ��

��

�

� � �

� �

ω

α

Let us assume the solution of Eq. (3.52) in the form

( ) i tZ f x e ω�

where f (x) satisfies

2

2

0

( )( ) 0

( )at 0

d f xi f x

dx

df xg x

dx

ωα

� �

� � �

(3.53)

(3.54)



Also,

( )f x is finite for large x. (3.55)

The solution of Eq. (3.53), satisfying the BC (3.55), is

( ) exp [ ( / ) ]f x A i xω α� �

The constant A can be determined by using the BC (3.54). Therefore,

01

( ) exp [ ( / ) ]f x g i xi

α ω αω

� �

Thus,

01

exp [ ( / ) ]Z g i t i xi

α ω ω αω

� � (3.56)

It can be shown for convenience that

1,

2

ii

�� 1 1

2

i

i

��

Thus, Eq. (3.56) can be written in the form

0

0

2exp (1 )exp

2 2 2

2exp (1 ) cos sin

2 2 2 2

gZ x i i t x

gx i t x i t x

α ω ωωω α α

α ω ω ωω ωω α α α

� ��

� ��

Its real part gives the fluctuation in temperature is

0

0

2( , ) exp cos sin

2 2 2 2

2exp cos

2 2 2 4

gT x t x t x t x

gx t x

α ω ω ωω ωω α α α

α ω ω πωω α α

��

� � � ��

Hence, the amplitude of the steady temperature is given by the factor

0 2exp

2 2

gx

α ωω α

� ��

EXAMPLE 3.9 Find the solution of the one-dimensional diffusion equation satisfying thefollowing BCs:

(i) T is bounded as t ��



(ii)0

0, for allx

Tt

x ��

�

(iii) 0, for allx a

Tt

x ��

�

(iv) ( , 0) ( ),T x x a x� � 0 .x a� �

Solution This is an example with insulated boundary conditions. From Section 3.5, itcan be seen that a physically acceptable general solution of the diffusion equation is

2( , ) exp ( ) ( cos sin )T x t t A x B xαλ λ λ� � �

Thus,

2exp ( ) ( sin cos )T

t A x B xx

��

� � � �αλ λ λ λ λ (3.57)

Using BC (ii), Eq. (3.57), gives B = 0. Since we are looking for a non-trivial solution, the useof BC (iii) into Eq. (3.57) at once gives

sin 0aλ � implying ,a nλ π� 0,1, 2,n � #

Using the principle of superposition, we get

22

0

( , ) exp ( ) cos exp cos .n nn

n nT x t A t x A t x

a a

π παλ λ α�

�

� ��

The boundary condition (iv) gives

2

01

( , 0) ( ) exp cosnn

n nT x x a x A A t x

a a

π πα�

�

� ��

where2

20

0

2

0

2 2

2 2 2 2

2( )

6

2( ) cos

2 2(1 cos ) [1 ( 1) ]

a

a

n

n

aA ax x dx

a

nA ax x x dx

a a

a an

n n

π

ππ π

� � �

� ��

� � �

�

�

Therefore,

2

2 2

4, for even

0, for odd

n

an

A n

n

π

��

� ��




22 2

2 22, 4, even

4 1( , ) cos exp

6 n

a a n nT x t x t

a an

π παπ

�

� �

� ��

EXAMPLE 3.10 The boundaries of the rectangle 0 , 0x a y b� � � � are maintained at zero

temperature. If at t = 0 the temperature T has the prescribed value f (x, y), show that fort > 0, the temperature at a point within the rectangle is given by

2

1 1

4( , , ) ( , ) exp ( ) sin sinmn

m n

m x n yT x y t f m n t

ab a b

π παλ� �

� ��

where

0 0( , ) ( , ) sin sin

a b m x n yf m n f x y dx dy

a b

π π� � �and

2 22 2

2 2mnm n

a bλ π

� ��

Solution The problem is to solve the diffusion equation described by

2 2

2 2PDE: , 0 , 0 , 0

BCs: (0, , ) ( , , ) 0, 0 , 0

( , 0, ) ( , , ) 0, 0 , 0

IC: ( , , 0) ( , ), 0 , 0

T T Tx a y b t

t x y

T y t T a y t y b t

T x t T x b t x a t

T x y f x y x a y b

α�

� � � � � � � ��

� � � � �

� � � � �

� � � � �

� � ��

Let the separable solution be

( ) ( ) ( )T X x Y y tβ�Substituting into PDE, we get

21X Y

X Y

�� β λα β

Then 2 0β αλ β� ��

2 2 (say)X Y

pX Y

�� λ



Hence,2 0X p X� ��

2 2 2 (say)Y

p qY

λ � � � � �

Therefore,2 0Y q Y� ��

Thus, the general solution of the given PDE is2

( , , ) ( cos sin ) ( cos sin ) tT x y t A px B px c qy D qy e αλ�� where

2 2 2p qλ � �

Using the BC: (0, , ) 0,T y t � we get A = 0. Then, the solution is of the form

2( , , ) sin ( cos sin ) tT x y t B px c qy D qy e αλ��

Applying the BC: ( , 0, ) 0,T x t � we get c = 0. Thus, the solution is given by

2( , , ) sin sin tT x y t BD px qye αλ��

Application of the BC: ( , , ) 0T a y t � gives

sin 0, implyingpa pa nπ� �

or

,n

pa

π� 1, 2,n � #

Using the principle of superposition, the solution can be written in the form

2

1

( , , ) sin sin tn

n

nT x y t A x qye

aαλπ��

�

� ��

Using the last BC: ( , , ) 0,T x b t � we obtain

,m

qb

π� 1, 2,m � #

Thus, the solution is found to be

2

1 1

( , , ) sin sin tmn

m n

n mT x y t A x y e

a b

� ��

� �

� � � �� αλπ π



where

2 22 2 2 2

2 2

m np q

b aλ π

� ��

Finally, using the IC, we get

( , , 0) ( , ) sin sinmnn m

T x y f x y A x ya b

� � � �� π π

which is a double Fourier series, where

0 0

2 2( , ) sin sin

a b

mnm n

A f x y x y dx dya b a b

� � � � � � � �� π π

Hence, the required general solution is

2

1 1

( , , ) ( , ) sin sint

m n

n mT x y t f m n e x y

a b

� ��

� �

� � � �� αλ π π

where

0 0, )

4( ( , ) sin sin

a bn

m nf m f x y x y dx dy

ab a b

π π= ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫and

2 22 2

2 2

m n

b aλ π

� ��

��

Consider a three-dimensional diffusion equation

2TT

t

��

� �α

In cylindrical coordinates ( , , ),r zθ it becomes

2 2 2

2 2 2 2

1 1 1T T T T T

t r rr r z

� � � � ��

� � � �α θ

(3.58)

where ( , , , ).T T r z t� θLet us assume separation of variables in the form

( , , , ) ( ) ( ) ( ) ( )T r z t R r H Z z t�θ θ β



Substituting into Eq. (3.58), it becomes

2

1 1R HZ R HZ H RZ Z RH RHZ

r r

�� ββ β β βα

or

22

1 1 1R R H Z

R r R H Zr

�� β λα β

where 2�λ is a separation constant. Then

2 0� ��β αλ β (3.59)

2 22

1 1(say)

R R H Z

R r R H Zr

�� λ μ

Thus, the equations determining Z, R and H become

2 0Z Z� �� μ (3.60)

2 22

1 10

R R H

R r R Hr

�� λ μ

or

2 2 2 2 2( ) (say)R R H

r r r vR R H

�� λ μ

Therefore,

2 0H v H� �� (3.61)

22 2

2

1( ) 0

vR R R

r r

��

�� λ μ (3.62)

Equations (3.59)–(3.61) have particular solutions of the form

2

cos sin

t

z z

e

H c v D v

Z Ae Be

�

�

�

� �

� �

αλ

μ μ

β

θ θ

The differential equation (3.62) is called Bessel’s equation of order v and its general solutionis known as

2 2 2 21 2( ) ( ) ( )v vR r c J r c Y r� � � �λ μ λ μ



where ( )J rν and ( )Y rν are Bessel functions of order ν of the first and second kind, respectively.

Of course, Eq. (3.62) is singular when r = 0. The physically meaningful solutions must be twice

continuously differentiable in 0 .r a� � Hence, Eq. (3.62) has only one bounded solution, i.e.

2 2( ) ( )R r J r� �ν λ μ

Finally, the general solution of Eq. (3.58) is given by

2 2 2( , , , ) [ ] [ cos sin ] ( )t z zT r z t e Ae Be C D J r� �� αλ μ μνθ νθ νθ λ μ

EXAMPLE 3.11 Determine the temperature ( , )T r t in the infinite cylinder 0 r a� � when

the initial temperature is ( , 0) ( ),T r f r� and the surface r = a is maintained at 0° temperature.

Solution The governing PDE from the data of the problem is

2TT

t

��

� �α

where T is a function of r and t only. Therefore,

2

2

1 1T T T

r r tr

� � ��

� �α

(3.63)

The corresponding boundary and initial conditions are given by

BC: ( , ) 0

IC: ( , 0) ( )

T a t

T r f r

�

�(3.64)

The general solution of Eq. (3.63) is

20( , ) exp ( ) ( )T r t A t J r� �αλ λ

Using the BC (3.64), we obtain

0 ( ) 0J a �λ

which has an infinite number of roots, ( 1, 2, , ).na n � � �ξ Thus, we get from the superposition

principle the equation

20

1

( , ) exp ( ) ( )n n nn

T r t A t J r�

�� αξ ξ

Now using the IC: ( , 0) ( ),T r f r� we get

01

( ) ( )n nn

f r A J r�

�� ξ



To compute An, we multiply both sides of the above equation by 0 ( )mrJ rξ and integrate with

respect to r to get

0 0 00 0

1

221

( ) ( ) ( ) ( )

0 for

( ) for2

a a

m n m nn

m m

rf r J r dr A rJ r J r dr

n m

aA J a n m

� � �

�

�

�

�

��

��

��

which gives

02 2 01

2( ) ( )

( )

a

m mm

A uf u J u dua J a

� � ξξ

Hence, the final solution of the problem is given by

2002 2 0

11

( )2( , ) exp ( ) ( ) ( )

( )

am

m mmm

J rT r t t uf u J u du

a J a

�

�

� �� ξ αξ ξξ

��

In this section, we shall examine the solution of diffusion or heat conduction equation in thespherical coordinate system. Let us consider the three-dimensional diffusion Eq. (3.9), and let

( , , , ).T T r t� θ φ In the spherical coordinate system, Eq. (3.9) can be written as

2 2

2 2 2 2 2

2 1 1 1sin

sin sin

T T T T T

r r tr r rθ

θ θ αθ θ φ� ��

� � � � � ��

(3.65)

This equation is separated by assuming the temperature function T in the form

( ) ( ) ( ) ( )T R r H t� θ φ βΦ (3.66)


22

2 2 2 2

2 1 1 1 1sin (say)

sin sin

R R d dH d

R r R H d dr r d

�� βθ λ

θ θ α βθ θ φΦ

Φ

where 2λ is a separation constant. Thus,

2 0d

dt� �β λ αβ



whose solution is2

1tc e�� αλβ (3.67)

Also,

2 22 2 2 2

2 2 2

1 2 1 1sin sin (say)

sin

d R dR d dH dr m

R r dr d ddr Hr d

� ��

Φθ θ λθ θ Φθ φ

which gives2

22

0d

mdφΦ Φ� �

whose solution is

1 2( ) im imc e c eφ φφΦ �� (3.68)

Now, the other separated equation is

2 22

2 2 2 2

1 2 1sin

sin sin

d R dR d dH m

R r dr d ddr Hr r

� � � �� θ λ

θ θθ θ

or

2 22 2

2

2 1sin

sinsin

( 1) (say)

r m d dHR R r

R r H d d

n n

� � � ��

� �

λ θθ θ θθ

On re-arrangement, this equation can be written as

22

2 ( 1)0

n nR R R

r r

�� λ (3.69)

and

2 2

2 2

1sin cos ( 1)

sin sin

d H dH mn n

H dd

� ��

θ θθ θθ θ

or

2 2

2 2cot ( 1) 0

sin

d H dH mn n H

dd

� �$ $� � � � �� $ $! "

θθθ θ

(3.70)

Let 1/2( ) ( );R r r�� λ ψ then Eq. (3.69) becomes

21/2 2

2

1 ( 1/2)) ( ) ( ) 0

nr r r

r r� ��

� �� λ ψ ψ λ ψ



Since ( ) 0,r %λ we have

22

2

1 ( 1/2)( ) ( ) ( ) 0

nr r r

r r

� ��$ $� � � �� $ $! "

ψ ψ λ ψ

which is Bessel’s differential equation of order ( 1/2),n � whose solution is

1/2 1/2( ) ( ) ( )n nr AJ r BY r� �� ψ λ λ

Therefore,

1/21/2 1/2( ) ( ) [ ( ) ( )]n nR r r AJ r BY r�� λ λ λ (3.71)

where Jn and Yn are Bessel functions of first and second kind, respectively. Now, Eq. (3.70)can be put in a more convenient form by introducing a new independent variable

cos�μ θ

so that

2

2

cot / 1

1dH dH

d d

� �

� � �

θ μ μ

μθ μ

2 22

2 2(1 )

d H d H dH

dd d� � �μ μ

μθ μThus, Eq. (3.70) becomes

2 22

2 2(1 ) 2 ( 1) 0

1

d H dH mn n H

dd

��

� �� μ μ

μμ μ(3.72)

which is an associated Legendre differential equation whose solution is

( ) ( ) ( )m mn nH A P B Q� �� θ μ μ (3.73)

where ( )mnP μ and ( )m

nQ μ are associated Legendre functions of degree n and of order m, of

first and second kind, respectively. Hence the physically meaningful general solution of thediffusion equation in spherical geometry is of the form

21/21/2

, ,

( , , , ) ( ) ( ) (cos )m im tmn n n

m n

T r t A r J r P e� � �� φ αλ

λλ

θ φ λ λ θ (3.74)

In this general solution, the functions ( )mnQ μ and 1/2

1/2( ) ( )nr Y r��λ λ are excluded because

these functions have poles at 1� &μ and r = 0 respectively.



EXAMPLE 3.12 Find the temperature in a sphere of radius a, when its surface is kept at

zero temperature and its initial temperature is ( , ).f r θ

Solution Here, the temperature is governed by the three-dimensional heat equation in

spherical polar coordinates independent of .φ Therefore, the task is to find the solution of

PDE

2

2 2

1 2 1sin

sin

T T T T

t r rr r

� � � � ��

�� θ

α θ θθ(3.75)

subject to

BC: ( , , ) 0

IC: ( , , 0) ( , )

T a t

T r f r

�

�

θ

θ θ

(3.76)

(3.77)

The general solution of Eq. (3.75), with the help of Eq. (3.74), can be written as

21/21/2

,

( , , ) ( ) ( ) (cos ) tn n n

n

T r t A r J r P e� �� αλ

λλ

θ λ λ θ (3.78)

Applying the BC (3.76), we get

1/2 ( ) 0nJ a� �λ

This equation has infinitely many positive roots. Denoting them by ,iξ we have

1/2 21/2

0 1

( , , ) ( ) ( ) (cos )exp ( )ni i n i n in i

T r t A r J r P t� �

��

� �� θ ξ ξ θ αξ (3.79)

Now, applying the IC and denoting cos by ,θ μ we get

1 1/21/2

0 1

( , cos ( )) ( ) ( ) ( ).ni i n i nn i

f r A r J r P� �

� ��

� �� μ ξ ξ μ

Multiplying both sides by ( )mP dμ μ and integrating between the limits, –1 to 1, we obtain

1 11 1/21/2

1 10 1

1/21/2

0 1

( , cos ( )) ( ) ( ) ( ) ( ) ( )

2( ) ( )

2 1

m ni i n i m nn i

ni i n in i

f r P d A r J r P P d

A r J rn

� ��

��

� ��

��

�

� ��

� �

� �

μ μ μ ξ ξ μ μ μ

ξ ξ



or

1 1 1/21/2

11

2 1( ) ( , cos ( )) ( ) ( )

2 n ni i n ii

nP f r d A r J r

��

��

�� μ μ μ ξ ξ for 0,1, 2, 3,n � #

Now, to evaluate the constants Ani, we multiply both sides of the above equation

by 3/21/2 ( )n jr J r� ξ and integrate with respect to r between the limits 0 to a and use the

orthogonality property of Bessel functions to get

11/2 3/2 11/2

0 1

1/2 1/20

1

21/2

1

2 1( ) ( ) ( , cos ( ))

2

( ) ( )

1[ ( )] , 0,1, 2, 3,

2

a

n j ni

a

ni n i n ji

ni n ii

nr J r dr P f r d

A rJ r J r dr

A J r n

��

�

� ��

�

��

� � ��

�

� � ��

� �

� �

�

ξ ξ μ μ μ

ξ ξ

ξ (3.80)

Thus, Eqs. (3.79) and (3.80) together constitutes the solution for the given problem.

�� !��

Theorem 3.1 (Maximum-minimum principle).

Let ( , )u x t be a continuous function and a solution of

t xxu u�α (3.81)

for 0 , 0 ,x l t T� � � � where T > 0 is a fixed time. Then the maximum and minimum values

of u are attained either at time t = 0 or at the end points x = 0 and x = l at some time in theinterval 0 .t T� �

Proof To start with, let us assume that the assertion is false. Let the maximum value

of ( , )u x t for 0(0 )t x l� � � or for x = 0 or x = l (0 )t T� � be denoted by M. We shall assume

that the function ( , )u x t attains its maximum at some interior point (x0, t0), in the rectangle

defined by 0 00 , 0 ,x l t T� � � � and then arrive at a contradiction. This means that

0 0( , )u x t M� � ε (3.82)

Now, we shall compare the signs in Eq. (3.81) at the point (x0, t0). It is well known fromcalculus that the necessary condition for the function u(x, t) to possess maximum at (x0, t0) is

0 0( , ) 0,u

x tx

��

2

0 02( , ) 0

ux t

x�

��

(3.83)



In addition, 0 0( , )u x t attains maximum for t = t0, implying

0 0( , ) 0u

x tt

��

(3.84)

Thus, with the help of Eqs. (3.83) and (3.84) we observe that the signs on the left- and right-hand sides of Eq. (3.81) are different. However, we cannot claim that we have reached acontradiction, since the left- and right-hand sides can simultaneously be zero.

To complete the proof, let us consider another point (x1, t1) at which 2 2/ 0u x� � � and

/ 0.u t� � �Now, we construct an auxiliary function

0( , ) ( , ) ( )v x t u x t t t� � �λ (3.85)

where λ is a constant. Obviously, 0 0 0 0( , ) ( , )v x t u x t M� � � ε and 0( ) .t t T� �λ λ Suppose

we choose 0,�λ such that /2 ;T�λ ε then the maximum of ( , )v x t for 0t � or

for 0,x x l� � cannot exceed the value /2.M � ε But ( , )v x t is a continuous function and,

therefore, a point 1 1( , )x t exists at which it assumes its maximum. It implies

1 1 0 0/2 ( , ) ( , )M v x t v x t M� � � � �ε ε

This pair of inequalities is inconsistent and therefore contradicts the assumption that v

takes on its maximum at 0 0( , ).x t Therefore, the assertion that u attains its maximum either

at 0t � or at the end points is true.We can establish a similar result for minimum values of u(x, t). If u satisfies Eq. (3.81),

–u also satisfies Eq. (3.81). Hence, both maximum and minimum values are attained eitherinitially or at the end points. Thus the proof is complete. We shall give some of the consequencesof the maximum-minimum principle in the following theorems.

Theorem 3.2 (Uniqueness theorem). Given a rectangular region defined by 0 , 0 ,x l t T� � � �and a continuous function u(x, t) defined on the boundary of the rectangle satisfying the heatequation

t xxu u�α

This equation possesses one and only one solution satisfying the initial and boundary conditions

1 2

( , 0) ( )

(0, ) ( ), ( , ) ( )

u x f x

u t g t u l t g t

�

� �

where f (x), g1(t), g2(t) are continuous on their domains of definition.

Proof Suppose there are two solutions 1 2( , ), ( , )u x t u x t satisfying the heat equation as

well as the same initial and boundary conditions. Now let us consider the difference



2 1( , ) ( , ) ( , )v x t u x t u x t= −

It is also a solution of the heat conduction equation for 0 , 0x l t T≤ ≤ ≤ ≤ and is continuous

in x and t. Also, ( , ) 0, 0v x t x l= ≤ ≤ and (0, ) ( , ) 0, 0 .v t v l t t T= = ≤ ≤ Hence, ( , )v x t satisfies

the conditions required for the application of maximum-minimum principle. Thus, ( , ) 0v x t = in

the rectangular region defined by 0 , 0 .x l t T≤ ≤ ≤ ≤ It follows therefore that 1 2( , ) ( , ).u x t u x t=Another important consequence of the maximum-minimum principle is the stability property

which is stated in the following theorem without proof.

Theorem 3.3 (Stability theorem). The solution ( , )u x t of the Dirichlet problem

, 0 , 0

( , 0) ( ), 0

(0, ) ( ), ( , ) ( ), 0

t xxu u x l t T

u x f x x l

u t g t u l t h t t T

= ≤ ≤ ≤ ≤

= ≤ ≤

= = ≤ ≤

α

depends continuously on the initial and boundary conditions.

3.9 NON-LINEAR EQUATIONS (MODELS)

Today various studies of fluid behaviour are available which encompass virtually any type ofphenomena of practical importance. However, there are many unresolved important problemsin fluid dynamics due to the non-linear nature of the governing PDEs and due to difficultiesencountered in many of the conventional, analytical and numerical techniques in solvingthem. In the following, we shall present few non-linear model equations to have a feel for thisvast field of study.

3.9.1 Semilinear Equations

Reaction–diffusion equations that appear in the literature are frequently semilinear and are ofthe form ut = —2u + f(u, x, t).

Typically, they appear as models in population dynamics, with inhomogeneous termdepending on the density of local population. In chemical engineering, f varies with temperatureand/or chemical concentration in a reaction like f(u) = luN.

3.9.2 Quasi-linear Equations

Many problems in fluid mechanics, when formulated mathematically, give rise to quasi-linearparabolic PDEs. A simple example concerns the flow of compressible fluid through a porousmedium. Let r denotes fluid density. Following Darcy’s law, which relates the velocity V tothe pressure p as

Kp

mÊ ˆ

= - —Á ˜Ë ¯V .



Then, the equation of conservation of mass is given by

( ) 0t

r r∂+ — ◊ =

∂V

and the equation of state p = p(r) can be combined to get

[ ( ) ]Kt

r r r∂ + — ◊ —∂

where K(r) is proportional to r dp

dr. The model can then be written as the porous medium

equation in the form

( )n

t

r r r∂+ — ◊ —

∂where n is a positive constant.

3.9.3 Burger’s Equation

The well-known Burger’s equation is non-linear and finding its solution has been the subjectof active research for many years. For simplicity, let us consider one-dimensional Burger’sequation in the form

ut + uux = vuxx

or21

02t x

x

u vu uÊ ˆ- - =Á ˜Ë ¯ (3.85a)

which is actually the non-linear momentum equation in fluid mechanics without the pressureterm. v is the physical viscosity. Here vuxx measures dissipative term and uux measuresconvective term, while ut is the unsteady term. Hopf (1950) and Cole (1951) gave independentlythe analytical solution for a model problem using a two-step Hopf–Cole transformation describedby

u(x, t) = yx

y = –2v log f(x, t)

That is,

2 xu vff

= - (3.85b)

Thus,

2

2 2xt x ttu v v

f f ff f

Ê ˆ= - +Á ˜Ë ¯

,



2

22 2xx x

xu v v

� ��

3

2 32 6 4xxx x xx x

xxu v v v

� ��

.

Inserting, these derivative expressions into Eq. (3.85a) and on simplification, we arrive at

x��

(vxx – t) – (vxx – t)x = 0. (3.85c)

Therefore, we have to solve Eq. (3.85c) to find (x, t), and using this result in Eq. (3.85b),we obtain an expression for u(x, t) which of course satisfies Eq. (3.85a). Thus, if (x, t)satisfies heat conduction equation

t = vxx (3.85d)

which also means solving trivially Eq. (3.85c). This is also called linearised Burger’s equation.Equivalently, we may introduce the transformation:

2

,

2

x

t x

u

uvu

�

�

� ��

(3.85e)

in such a way, satisfying that �xt = �tx. Then, the above transformation can be rewritten as

�t = v�xx – 2

2x�

(3.85f)

Also, Eq. (3.85b) can be recast in the form

(x, t) = e[–�(x, t)/2v] (3.85g)

Thus, knowing (x, t), we can find u(x, t) from Eq. (3.85b). It may also be observed thatEqs. (3.85d) and (3.85f) are equivalent.

Hence, the transformation of non-linear Burger’s equation into heat conduction equation,made life easy to get analytical solution to the Burger’s equation.

��"�� )'.'*(��*(-%� �+/3(%&� 4/+��-+1%+20� ,-*.'/)

The IVP for Burger’s equation can be stated as follows. Solve

PDE: ut + uux = vuxx, –� < x < �, t > 0

IC: u(x, 0) = f(x) (3.85h)

Under the transformation defined by Eq. (3.85b) and using (3.85g), the given IVP can berestated as a Cauchy problem, described by

PDE: t = vxx,



IC: (x, 0) = 0

1( )

2( )

xf d

vx e� �

��

(3.85i)

Using, the standard, separation of variables, method of solution, as given in Eq. (3.28), thesolution to Eq. (3.85i) is found to be

21 ( )( , ) ( )exp

44

xx t x d

vtvt

� �

�

��

� ��

� � � (3.85j)

Substituting the expression for ( )x� , the above equation can be rewritten as

1 ( , , )

( , ) exp24

f x tx t d

vvt

��

�

��

� �� where,

2

0

( )( , , ) ( )

2

xf x t f d

t

� �� .

Finally, using Eq. (3.85b), the exact solution of the IVP for Burger’s equation as statedin Eq. (3.85h) is found to be

( ) ( , , )exp

2( , )

( , , )exp

2

x f x td

t vu x t

f x td

v

� � �

� �

�

��

� ��

�(3.85k)

Here, the function f(�, x, t) is known as Hopf–Cole function.

��5 ��

EXAMPLE 3.13 A homogeneous solid sphere of radius R has the initial temperature distribution

( ), 0 ,f r r R� � where r is the distance measured from the centre. The surface temperature

is maintained at 0°. Show that the temperature ( , )T r t in the sphere is the solution of

2 2t rr rT c T T

r� � ��

where c2 is a constant. Show also that the temperature in the sphere for t > 0 is given by

2

1

, )1

( sin exp ( ),n nn

tn

T r B r tr R

�

�� π λ

ncn

R� πλ



Solution The temperature distribution in a solid sphere is governed by the parabolicheat equation

2 2tT c T� �

From the data given, T is a function of r and t alone. In view of the symmetry of the sphere,the above equation with the help of Eq. (3.65) reduces to

2 2t rr rT c T T

r� ��

(3.86)

Setting ,v rT� the given BC gives

( , ) ( , ) 0v R t rT R t� �

while the IC gives

( , 0) ( , 0) ( )v r rT r rf r� �

Since T must be bounded at 0,r � we require

2

(0, ) 0

Now,

, rt t r

r

v t

v r vvv rT T

r r

�

��

Similarly, finding Trr and substituting into Eq. (3.86), we obtain2

t rrv c v�

Using the variables separable method, we may write ( , ) ( ) ( )v r t R r t� τ and get

2 2

( ) cos sin

( ) exp ( )

R r A kr B kr

t c k t

� �

� �τ

Thus, using the principle of superposition, we get

2 2

1

( , ) ( cos sin ) exp ( )n nn

v r t A kr B kr c k t�

��

Also, using (0, ) 0,v t � we have

0( cos sin ) | 0n n rA kr B kr ��

implying 0.nA � Also, ( , ) 0v R t � gives Bn sin kR = 0, implying sin kR = 0, as 0.nB Therefore,

,kR n� π ,n

kR

� π1, 2,n � #



Thus, the possible solution is

2 2 2

21

( , ) sin expnn

n c n tv r t B r

R R

π π�

�

� ��

Finally, applying the IC: ( , 0) ( ),v r rf r� we get

1

( ) sinnn

nrf r B r

R

�

�

� �� π

which is a half-range Fourier series. Therefore,

0

2( )sin

R

nn

B rf r r drR R

� �� π

But ( , ) ( , ).v r t rT r t� Hence, the temperature in the sphere is given by

2 2 2

21

1( , ) sin expn

n

n c n tT r t B r

r R R

�

�

� �� π π

EXAMPLE 3.14 A circular cylinder of radius a has its surface kept at a constant temperatureT0. If the initial temperature is zero throughout the cylinder, prove that for t > 0.

200

11

( )2( , ) 1 exp ( )

( )n

nn nn

J aT r t T kt

a J a

�

�

� ��

� ξ ξξ ξ

where 1 2, , , n� � �ξ ξ ξ are the roots of 0 ( ) 0,J a �ξ and k is the thermal conductivity which

is a constant.

Solution It is evident that T is a function of r and t alone and, therefore, the PDE tobe solved is

2

2

1 1T T T

r r k tr

� � ��

� � (3.87)

subject to

0

IC : ( , 0) 0, 0

BC : ( , ) , 0

T r r a

T a t T t

� � �

�

Let

0 1( , ) ( , )T r t T T r t� �



so that

1 0

1

( , 0)

( , ) 0

T r T

T a t

� �

�

(3.88)

(3.89)

where T1 is the solution of Eq. (3.87). By the variables separable method we have (seeExample 3.11),

21 0( , ) ( ) exp ( )T r t AJ r kt� �λ λ

Using the BC: 1( , ) 0,T a t � we get

20 ( ) exp( ) 0AJ a kt� �λ λ

which gives 0 ( ) 0 0.J a as A� λ Let 1 2, , , ,nξ ξ ξ be the roots of 0 ( ) 0.J a �λ Then the

possible solution using the superposition principle is

21 0

1

( , ) ( ) exp ( )n n nn

T r t A J r kt�

�� ξ ξ (3.90)

Using the IC: 1 0( , 0)T r T� � into Eq. (3.90), we obtain

0 01

( )n nn

A J r T�

�� ξ

Multiplying both sides by 0 ( )mrJ rξ and integrating, we get

0 0 0 00 0

1

20

0

221

( ) ( ) ( )

( ) if ; otherwise 0

( )2

a a

m n m nn

a

m m

m m

T rJ r dr A rJ r J r dr

A rJ r m n

aA J a

�

��

� �

�

��

�

ξ ξ ξ

ξ

ξ

But,

0 0 0 00 0

012 0

0 01 102

( ) ( ) ( )

[ ( )]

[ ( )] ( )

m

m

m

a a

m mm m

a

m

am

mm

x dxT rJ r dr T J x x r

T dxJ x dx

dx

T aTxJ x J a

� � � �

� �

� � � �

� �

�

ξ

ξ

ξ

ξ ξξ ξ

ξ

ξξξ



Therefore,

22 01 1( ) ( )

2m m mm

aTaA J a J a� �ξ ξ

ξor

0

1

2 1

( )� �� n

n n

TA

a J a

Hence, Eq. (3.90) becomes

20

1 011

exp ( )( )2( , )

( )nn

n nn

ktJ rT r t T

a J a

�

�

��

ξξξ ξ

Finally, the complete solution is found to be

20

011

exp ( )( )2( , ) 1

( )nn

n nn

ktJ rT r t T

a J a

�

�

� ��

�ξξ

ξ ξ

EXAMPLE 3.15 Determine the temperature in a sphere of radius a, when its surface is

maintained at zero temperature while its initial temperature is ( , ).f r θ

Solution Here the temperature is governed by the three-dimensional heat equation in

polar coordinates independent of ,φ which is given by

2

2 2

1 2 1sin

sin

u u u u

k t r rr r

� � � � ��

�� θ

θ θθ(3.91)

Let

( , , ) ( ) ( ) ( )u r t R r H T t�θ θ

By the variables separable method (see Section 3.7), the general solution of Eq. (3.91) isfound to be

1/2 21/2( , , ) ( ) ( ) (cos ) exp ( )n n n

n

u r t A r J r P k t�� λ

λθ λ λ θ λ (3.92)

In the present problem, the boundary and initial conditions are

BC: ( , , ) 0

IC: ( , , 0) ( , )

u a t

u r f r

�

�

θ

θ θ

(3.93)

(3.94)

Substituting the BC (3.93) into Eq. (3.92), we get

1/2 ( ) 0nJ a� �λ (3.95)



Let 1 2, , , ,ia a a… …ξ ξ ξ be the roots of Eq. (3.95). Then the general solution can be put in the form

1/2 21/2

1 1( , , ) ( ) ( ) (cos )exp ( )ni i n i n i

n iu r t A r J r P k t

∞ ∞−

+= =

= −∑ ∑θ ξ ξ θ ξ (3.96)

Now using the IC, we obtain

1/21/2

1 1( , ) ( ) ( ) (cos )ni i n i n

n if r A r J r P

∞ ∞−

+= =

= ∑ ∑θ ξ ξ θ

Multiplying both sides by (cos ) (cos )nP dθ θ and integrating, we have

1 11/2 21/21 11 1

(cos ) ( , ) (cos ) ( ) ( ) (cos ) (cos )n ni i n i nn i

P f r d A r J r P d∞ ∞

−+− −= =

= ∑ ∑∫ ∫θ θ θ ξ ξ θ θ

Using the orthogonality property of Legendre polynomials, we get

1 1/21/21 1 1

2(cos ) ( , ) (cos ) ( ) ( )2 1n ni i n i

n iP f r d A r J r

nθ θ θ ξ ξ

∞ ∞−

+− = =

⎛ ⎞= ⎜ ⎟⎝ ⎠+∑ ∑∫

Rearranging and multiplying both sides of the above equation by 3/21/2 ( )n ir J r+ ξ and integrating

between the limits 0 to a with respect to r, we get1 2 1/23/2

1/2 1/20 1 0

22

1/2

2 1 ( ) (cos ) ( , ) (cos ) ( )2

{ ( )}2

a an i n ni i in

ni n i

n r J r dr P f r d A rJ r dr

aA J a

−+ +−

+

+ =

= ′

∫ ∫ ∫ξ θ θ θ ξ ξ

ξ

Therefore,

1/213/2

1/22 2 0 11/2

(2 1)( ) (cos ) ( , ) (cos )

{ ( )}

aini n i n

n i

nA r J r dr P f r d

a J a + −+

+=

′ ∫ ∫ξ

ξ θ θ θξ

(3.97)

Hence, we obtain the solution to the given problem from Eq. (3.96), where Ani is given byEq. (3.97).

EXAMPLE 3.16 The heat conduction in a thin round insulated rod with heat sources presentis described by the PDE

( , )/ ,t xxu u F x t cα ρ− = 0 , 0x l t< < > (3.98)

subject to

BCs: (0, ) ( , ) 0

IC: ( , 0) ( ), 0

u t u l t

u x f x x l

= =

= ≤ ≤ (3.99)

where ρ and c are constants and F is a continuous function of x and t. Find ( , ).u x t



Solution It can be noted that the boundary conditions are of homogeneous type. Letus consider the homogeneous equation

0t xxu u� �α (3.100)

Setting ( , ) ( ) ( ),u x t X x T t� we get

2 (say)T X

T Xλ

α� �� (3.101)

which gives 2 0.X Xλ� �� The corresponding BCs are

(0) ( ) 0X X l� �

The solution of Eq. (3.101) gives the desired eigenfunctions and eigenvalues, which are

( ) sin ,n nX x x� λ2

2 ,nn

l� �� πλ 1n � (3.102)

For the non-homogeneous problem (3.98), let us propose a solution of the form

1

( , ) ( ) ( )n nn

u x t T t X x�

�� (3.103)

It is clear that Eq. (3.103) satisfies the BCs (3.99). From the orthogonality of eigenfunctions,it follows that

0

2( ) ( , ) ( )

l

m mT t u x t X x dxl

� �However,

1

0

2(0) ( ) sinm

mT f x x dx

l l� � � �� π

(3.104)

which is an IC for Tm(t). Introducing Eq. (3.103) into the governing equation (3.98), we get

1 1

( , )n n n n

n n

F x tT X T X

c

� �

� �� α ρ

(3.105)

Now, we shall expand ( , )/ ,F x t cρ so that it is represented by a convergent series

on 0 , 0x l t� � in the form

1

( ) ( )n nn

Fq t X x

c

�

�� ρ

(3.106)



where

0

2 ( , )( ) sin

l

nF x t n

q t x dxl c l

� �� πρ

(3.107)

Thus, qn(t) is known. Now, Eq. (3.105), with the help of Eq. (3.101), becomes

2

1

( ) 0n n n n nn

X T T q�

�� λ α

Therefore, it follows that

2( ) ( ) ( )n n n nT t T t q t� �� λ α (3.108)

Its solution with the help of IC (3.104) is

2 2

0( ) (0) exp ( ) exp [ ( )] ( )

t

n n n n nT t T t t q d� � � ��λ α λ α τ τ τ (3.109)

From Eqs. (3.103) and (3.109), the complete solution is found to be

12 2

01

( , ) (0) exp ( ) exp [ ( ) ( ) ( )n n n n nn

u x t T t t q d X x�

�

�� λ α λ α τ τ τ

In the expanded form, it becomes

� �

2

01

2

0 0

2( , ) ( ) ( ) exp ( )

2 ( , )exp ( ) ( ) ( )

l

n nn

l l

n n n

u x t f X d tl

Ft X d d X x

l c

�

�

��

��

�

� �

� �

ξ ξ ξ λ α

ξ τλ α τ ξ ξ τρ

(3.110)

It can be verified that the series in Eq. (3.110) converges uniformly for t > 0. By changingthe order of integration and summation in Eq. (3.110), we get

2

01

2

0 01

exp ( ) ( ) ( )( , ) ( )

1/2

exp { ( )} ( ) ( ) ( , )

/2

l n n n

n

l l n n n

n

t X x Xu x t f d

t X x X F td d

l c

�

�

�

�

� ��

� ��

��

��

λ α ξξ ξ

λ α τ ξ ξ ξ τρ

which can also be written in the form

0 0 0

( , )( , ) ( , ; ) ( ) ( , ; )

l l l Fu x t G x t f d G x t d d

c� � �� ξ τξ ξ ξ ξ τ ξ τ

ρ (3.111)



where

2

1

exp ( ) ( ) ( )( , ; )

/2n n n

n

t X x XG x t

l

�

�

��

λ α ξξ

is called Green’s function. More details on Green’s function are given in Chapter 5.

EXAMPLE 3.17 The temperature distribution of a homogeneous thin rod, whose surface isinsulated is described by the following IBVP:

PDE: vt – vxx = 0, 0 < x < L, 0 < t < � (3.112)

BCS: v(0, t) = v(L, t) = 0 (3.113)

IC: v(x, 0) = f(x), 0 � x � L (3.114)

Find its formal solution.

Solution Let us assume the solution in the form

v(x, t) = X(x)T(t)

Eq. (3.112) gives

XT� = X�T

or 2 (say)X T

X T��

where � is a positive constant. Then, we have

X� + �2X = 0

and T� + �2T = 0

From the BCS

v(0, t) = X(0)T(t) = 0,

and v(L, t) = X(L)T(t) = 0,

we obtain, X(0) = X(L) = 0 for arbitrary t. Thus, we have to solve the eigenvalue problem

X� + �2X = 0

subject to X(0) = X(L) = 0.The solution of the differential equation is

X(x) = A cos �x + B sin �x.

Since X(0) = 0, A = 0. The second condition yields

X(L) = B sin �L = 0

For non-trivial solution, B � 0 and therefore we have

sin �L = 0, implying � = n�/L, for n = 1, 2, 3, …



Thus, the solution is obtained as

Xn(x) = Bn sin n x

L

p .

Next, we consider the equation

T¢ + a2T = 0

whose solution can be written as

2

( ) tT t Ce a-=

or2( / )( ) n L t

n nT t C e p-= .

Hence, the non-trivial solution of the given heat equation satisfying both the boundaryconditions is found to be

2( / )( , ) sinn L tn n

n xv x t a e

Lp p- Ê ˆ= Á ˜Ë ¯

(3.115)

where an = bncn (arbitrary constant).To satisfy the IC, we should have

1

( , 0) ( ) sinnn

n xv x f x a

L

p•

=

Ê ˆ= = Á ˜Ë ¯Âwhich holds good, if f(x) is representable as Fourier Sine series with Fourier coefficients

0

2( )sin

L

nn x

a f x dxL L

pÊ ˆ= Á ˜Ë ¯Ú .

Hence, the required formal solution is

01

2( , ) ( )sin

L

n

nv x t f d

L L

ptt t•

=

È ˘= Í ˙Î ˚Â Ú p p- Ê ˆÁ ˜Ë ¯

2( / ) sinn L t n xe

L.

EXERCISES

1. A conducting bar of uniform cross-section lies along the x-axis, with its ends at x = 0and x = l. The lateral surface is insulated. There are no heat sources within the body.

The ends are also insulated. The initial temperature is 2 , 0 .lx x x l− ≤ ≤ Find thetemperature distribution in the bar for t > 0.

2. The faces 0,x x a= = of a finite slab are maintained at zero temperature. The initial

distribution of temperature in the slab is given by ( , 0) ( ), 0 .T x f x x a= ≤ ≤ Determinethe temperature at subsequent times.



3. Show that the solution of the equation

2

2

T T

t x

� ��

�

satisfying the conditions:

(i) 0 asT t� � �(ii) T = 0 for x = 0 and x = a for all t > 0

(iii) T = x when t = 0 and 0 < x < a

is

12

1

, )2 ( 1)

( sin exp [ ( / ) ]n

n

ta n

T x x n a tn a

� �

��

� � � � � �� π ππ

4. Solve the equation2

2

T T

t x

� ��

� satisfying the conditions:

(i) T = 0 when x = 0 and 1 for all t

(ii)

2 , 0 1/2

12(1 ), 1 when 0.

2

x x

Tx x t

� ��$� �

� � � �$!

5. Solve the diffusion equation

2

2

1v

t r rr

θ θ θ� ��

� � ��

subject to

2 2

0, is finite, 0

, 0, 0

( ), 04

r t

r a t

Pa r t

� �

� � �

� � �

θ

θ

θμ

Here, P, andμ ν are constants.

6. A homogeneous solid sphere of radius R has the initial temperaturedistribution ( ), 0 ,f r r R� � where r is the distance measured from the centre. Thesurface temperature is maintained at 0°. Show that the temperature T(r, t) in the

sphere is the solution of 2 2.t rr rT c T T

r� � ��

Show that the temperature in the sphere

for t > 0 is given by



2

1

1( , ) sin exp ( )n n

n

nT r t B r t

r R

�

�

� �� π λ

where /n cn R�λ π and c2 is a constant.

7. If ( )xφ is bounded for all real values of x, show that

21( , ) ( ) exp [ ( ) /(4 )]

4T x t x kt d

kt

�

�� φ ξ ξ ξ

π

is a solution of t xxT kT� such that ( , 0) ( ).T x x� φ8. An infinite homogeneous solid circular cylinder of radius a is thermally insulated to

prevent heat escape. At any time t, the temperature T(r, t) at a distance r from theaxis of symmetry is given by the heat conduction equation with axial symmetry. Attime t = 0, the initial temperature distribution at a distance r from the axis is knownto be a function of r. Find the temperature distribution at any subsequent time.

9. Let ( , , )r x y z� represent a point in three-dimensional Euclidean space R3. Find a

formal solution ( , )u r t which satisfies the diffusion equation

2 ,tu u� �α 0t �

and the BC: ( , 0) ( ),u r f r� where 3.r R�

10. Solve2

2, 0 , 0x a t

t x

� ��

� � � �θ θsubject to the conditions

0(0, ) ( , ) 0 and ( , 0) (constant).t a t x� � �θ θ θ θ

(GATE-Maths, 1996)


��

One of the most important and typical homogeneous hyperbolic differential equations is thewave equation. It is of the form

22 2

2

uc u

t� �

��

��

where c is the wave speed. This differential equation is used in many branches of Physics andEngineering and is seen in many situations such as transverse vibrations of a string ormembrane, longitudinal vibrations in a bar, propagation of sound waves, electromagneticwaves, sea waves, elastic waves in solids, and surface waves as in earthquakes. The solutionof a wave equation is called a wave function.

An example for inhomogeneous wave equation is

22 2

2

uc u F

t� � �

��

��

where F is a given function of spatial variables and time. In physical problems F representsan external driving force such as gravity force. Another related equation is

22 2

22

u uc u F

ttγ� � � �

� ��

(4.3)

where γ is a real positive constant. This equation is called a wave equation with dampingterm, the amplitude of which decreases exponentially as t increases. In Section 4.2, we shallderive the partial differential equation describing the transverse vibration of a string.

232

��

Hyperbolic Differential Equations


HYPERBOLIC DIFFERENTIAL EQUATIONS 233

��

Suppose a flexible string is stretched under tension τ between two points at a distance Lapart as shown in Fig. 4.1. We assume the following:

1. The motion takes place in one plane only and in this plane each particle moves ina direction perpendicular to the equilibrium position of the string.

2. The tension τ in the string is constant.

τ

τP

Qds

dx A

y y x t = ( , )

Ox

y

Fig. 4.1 Flexible string.

3. The gravitational force is neglected as compared with tension τ of the string.4. The slope of the deflection curve is small.

Let the two fixed ends of the string be at the origin O and A(L, 0) which lies along thex-axis in its equilibrium position. Consider an infinitesimal segment PQ of the string. Let ρ bethe mass per unit length of the string. If the string is set vibrating in the xy-plane, thesubsequent displacement, y from the equilibrium position of a point P of the string will bea function of x and time t, while an element of length dx is stretched into an element of lengthds given by

2 211 1

2

y yds dx dx

x x

� ��

� ��

The elementary elongation is given by

21

2

ydL ds dx dx

x� ��

while the work done by this element against the tension τ is

21

2

ydx

x��

Therefore, the total work done, W, for the whole string is

2

0

1

2

L yW dx

x��



If U is the potential energy of the string, then

2

0

12

L yU W dxx

τ ⎛ ⎞= = ⎜ ⎟⎝ ⎠∫ ∂∂

Also, the total kinetic energy T of the string is given by

2

0

12

L yT dxt

ρ ⎛ ⎞= ⎜ ⎟⎝ ⎠∫ ∂∂

Using Hamilton’s principle (See—Sankara Rao, 2005), we have

1

0( ) 0

t

tT U dt− =∫δ

i.e.

1

0( )

t

tT U dt−∫

is stationary. In other words,

1

0

2 2

0

12

t L

t

y y dx dtt x

ρ τ⎡ ⎤⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

∫ ∫ ∂ ∂∂ ∂

is stationary, and is of the form

( , , , , )x tF x t y y y dx dt∫∫Noting that x and t are independent variables, from the Euler-Ostrogradsky equation, we have

0t x

F F Fy t y x y

⎛ ⎞ ⎛ ⎞− − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∂ ∂ ∂ ∂ ∂∂ ∂ ∂ ∂ ∂

which gives

0y yt t x x

ρ τ⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∂ ∂ ∂ ∂∂ ∂ ∂ ∂

If the string is homogeneous, then ρ and τ are constants, in which case the governing equationrepresenting the transverse vibration of a string is given by

2 22

2 2y yc

t x=

∂ ∂∂ ∂

(4.4)

where

2 /c = τ ρ (4.5)



EXAMPLE 4.1 Consider Maxwell’s equations of electromagnetic theory given by

4

0

1

4 1

c t

i

c c t

πρ

π

E

H

HE

EH

� � �

� � �

� � � �

� � � �

��

��

where E is an electric field, ρ is electric charge density, H is the magnetic field, i is thecurrent density, and c is the velocity of light. Show that in the absence of charges, i.e.,

when 0,i� �ρ E and H satisfy the wave equations.

Solution Given

1curl

c t

HE E� � � � � �

�

Taking its curl again, we get

2

2 2

1 1( )

1 1 1

c t t c

c t c t c t

HE H

E E

� � � ��

� ��

� ��

� � ��

Moreover, using the identity

2 2( ) ( )E E E E� � � � � � � � � � � ��

it follows directly that

22

2 2

1

c t

EE� �

��

Similarly, we can observe that the magnetic field H also satisfies

22

2 2

1

c t

HH� � �

�

which is also a wave equation.



��

The one-dimensional wave equation is

2 0tt xxu c u� � (4.6)

Choosing the characteristic lines

,x ct x ct� �� (4.7)

the chain rule of partial differentiation gives

( )

x x x

t t t

u u u u u

u u u c u u

ξ η ξ η

ξ η η ξ

ξ η

ξ η

� � � �

� � � �

In the operator notation we have

, cx t

� � � � � ��

� � � � � ��

Thus, we get

22

22

uu u u u

x��

� � � � � � ��

� � ��

(4.8)

22

2( 2 )

uc u u u

t��

�(4.9)

Substituting Eqs. (4.8) and (4.9) into Eq. (4.6), we obtain

4 0u�� (4.10)

Integrating, we get

( , ) ( ) ( ),u � � � � � ��

where � and � are arbitrary functions. Replacing � and � as defined in Eq. (4.7), we have

the general solution of the wave equation (4.6) in the form

( , ) ( ) ( )u x t x ct x ct� �� (4.11)

The two terms in Eq. (4.11) can be interpreted as waves travelling to the right and left,respectively. Consider

1( , ) ( )u x t x ct��



This represents a wave travelling to the right with speed c whose shape does not change as

it travels, the initial shape being given by a known function ( ).xφ In fact, by setting 0t =in the argument of ,φ it can be observed that the initial wave profile is given by

1( , 0) ( )u x xφ=

At time 1/ ,t c=

1( , 1/ ) ( 1)u x c xφ= −

Let 1.x x= −′ Then ( 1) ( ).x xφ φ− = ′ That is, the same shape is retained even if the origin

is shifted by one unit along the x-axis. In other words, the graph of 1( ,1/ )u x c is the same as

the graph of the original wave profile translated one unit to the right. At 2/ ,t c= the graph

of 1( , 2 / )u x c is the graph of the wave profile translated two units to the right. Thus, in

particular, at 1,t = we have 1( ,1) ( ).u x x cφ= − Hence in one unit of time, the profile has

moved c units to the right. Therefore, c is the wave speed or speed of propagation. Using

similar argument, we can conclude that the equation 2 ( , ) ( )u x t x ctψ= + is also a wave profile

travelling to the left with speed c along the x-axis. Hence the general solution (4.11) of one-dimensional wave equation represents the superposition of two arbitrary wave profiles, bothof which are travelling with a common speed but in the opposite directions along the x-axis,while their forms remain unaltered as they travel. This situation is described in Fig. 4.2.

0–1 (1, 1/ )c ( , 1)c

u x x1( , 0) = ( )φ

u x c1( , 1/ ) u x1( , 1)

x

u

Fig. 4.2 Travelling wave profile.

Let k be an arbitrary real parameter. Consider then

( , ) [ ( )] [ ( )]u x t k x ct k x ctφ ψ= − + + (4.12)

This is also a solution of the one-dimensional wave equation. Further, let .kcω = Then

( , ) [ ] [ ]u x t kx t kx tφ ω ψ ω= − + + (4.13)



A function of the type given in Eq. (4.13) is a solution of one-dimensional wave equation iff

.kcω = Therefore, waves travelling with speeds which are not the same as c cannot be

described by the solution of the wave equation (4.6). Here, ( )kx tω+ is called the phase for

the left travelling wave. We have already noted that x ct± are the characteristics of the one-dimensional wave equation.

EXAMPLE 4.2 Obtain the periodic solution of the wave equation in the form( )( , ) i kx tu x t Ae ω±=

where 1, / ,i k cω= − = ± A is constant; and hence define various terms involved in wave

propagation.

Solution Let ( , ) ( ) i tu x t f x e ω±= be a solution of the wave equation

2tt xxu c u=

Then

2( ) , ( )w ww± ±= = -¢¢ i t i txx ttu f x e u f x e

Substituting into the wave equation, we get

2

2( ) ( ) 0f x f xcω

+ =′′

Its general solution is found to be

1 2( ) exp [ ( / ) ] exp [ ( / ) ]f x c i c x c i c xω ω= + −

Therefore, the required solution of the wave equation is

1 2( , ) [ exp { ( / ) } exp { ( / ) }] i tu x t c i c x c i c x e ωω ω ±= + −

Since / ,k cω= ± the time-dependent wave functions are of the form

( )( , ) i kx tu x t Ae ω±=

Hence, ( )( , ) i kx tu x t Ae ω±= is a solution of the wave equation, and is called a wave function.

It is also called a plane harmonic wave or monochromatic wave. Here, A is called the

amplitude, ω the angular or circular frequency, and k is the wave number, defined as the



number of waves per unit distance. By taking the real and imaginary parts of the solution,we find the linear combination of terms of the form

cos ( ), sin ( )A kx t A kx tω ω± ±

representing periodic plane waves. For instance, consider the function ( , ) sin ( ).u x t A kx tω= −This is a sinusoidal wave profile moving towards the right along the x-axis with speed .cDefining the wave length λ as the length over which one full cycle is completed, we

have / ,kλ π= 2 thereby implying that 2 / .k π λ=

Suppose an observer is stationed at a fixed point 0;x then,

0 0

0 0

, sin

sin ( 2 ) sin ( )

u x t A kx tc c

A kx t A kx t

λ λω ω

ω π ω

⎛ ⎞ ⎛ ⎞+ = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − − = −

Thus, we have

0 0( , / ) ( , )u x t c u x tλ+ =

Hence, exactly one complete wave passes the observer in time / ,T cλ= which is called the

period of the wave. The reciprocal of the period is called frequency and is denoted by

1/f T=

The function, cos ( ) sin ( /2 ),u A kx t A kx tω π ω= − = + − also represents a wave train except

that it differs in phase by /2π from the sinusoidal wave. Now consider the superposition of

the sinusoidal waves having the same amplitude, speed, frequency, but moving in oppositedirections. Thus, we have

( , ) sin [ ( )] sin [ ( )]

2 sin cos ( ) 2 cos ( ) sin

u x t A k x ct A k x ct

A kx kct A kct kx

= − + +

= =

Its amplitude factor [2 cos ( )]A kct varies sinusoidally with frequency .ω This situation is

described as a standing wave. The points / , 0, 1, 2,nx n k nπ= = ± ± … are called nodes. No

displacement takes place at a node. Therefore,

( , ) 0 for allnu x t t=

The nth standing wave profile will have ( 1)n − equally spaced nodes in a given interval

as shown in Fig. 4.3.



n = 5

n = 4

n = 3

n = 2

n = 1

Fig. 4.3 Standing wave profiles.

4.4 THE INITIAL VALUE PROBLEM; D’ALEMBERT’S SOLUTION

Consider the initial value problem of Cauchy type described as

2PDE: 0, , 0tt xxu c u x t− = −∞ < < ∞ ≥ (4.14)

ICs: ( , 0) ( ), ( , 0) ( )tu x x u x v xη= = (4.15)

where the curve on which the initial data ( )xη and ( )v x are prescribed is the x-axis. ( ) and ( )x v xηare assumed to be twice continuously differentiable. Here, the string considered is of an

infinite extent. Let ( , )u x t denote the displacement for any x and t. At 0,t = let the displacement

and velocity of the string be prescribed. We have already noted in Section 4.3 that the generalsolution of the wave equation is given by

( , ) ( ) ( )u x t x ct x ctφ ψ= + + − (4.16)

where φ and ψ are arbitrary functions. Substituting the ICs (4.15) into Eq. (4.16), we obtain

( ) ( ) ( )

[ ( ) ( )] ( )

x x x

c x x v x

φ ψ η

φ ψ

+ =

′ ′− =(4.17)

Integrating the second equation of (4.17), we have

0

1( ) ( ) ( )x

x x v dc

φ ψ ξ ξ− = ∫



Addition and substraction of this equation with the first relation of Eqs. (4.17) yield

0

( ) 1( ) ( )

2 2

xxx v d

c

ηφ ξ ξ= + ∫

0

( ) 1( ) ( )

2 2

xxx v d

c

ηψ ξ ξ= − ∫respectively. Substituting these relations for ( ) and ( )x x� � into Eq. (4.16), we at once have

1 1( , ) [ ( ) ( )] ( )

2 2

x ct

x ctu x t x ct x ct v d

cη η ξ ξ

+

−= + + − + ∫ (4.18)

This is known as the D’Alembert’s solution of the one-dimensional wave equation. If = 0,v i.e.,if the string is released from rest, the required solution is

1( , ) [ ( ) ( )]

2u x t x ct x ct� �� (4.19)

The D’Alembert’s solution has an interesting interpretation as given in Fig. 4.4.

x

c2

c1

c3

A B

P x t( , )0 0

xct

xct

–

=

–

0

0 xct

xct

+ =

+ 0

0

( + x ct0 0, 0)( – x ct0 0, 0)O

t

IR

Fig. 4.4 Characteristic triangle.

Consider the xt-plane and a point 0 0( , ).P x t Draw two characteristics through P backwards,

until they intersect the initial line, i.e., the x-axis at and .A B The equation of these two

characteristics are

0 0x ct x ct� � �

Equation (4.18) reveals that the solution 0 0( , ) at ( , )u x t P x t can be obtained by averaging the

value of � at andA B and integrating v along the x-axis between and .A B Thus, to find the

solution of the wave equation at a given point P in the xt-plane, we should know the initialdata on the segment AB of the initial line which is obtained by drawing the characteristicsbackward from P to the initial line. Here the segment AB of the initial line, on which the value



of ( , ) atu x t P depends, is called the domain of dependence of ,P and the triangle PAB is

called the characteristic triangle (see Fig. 4.4), which is also called the domain of determinancyof the interval.

EXAMPLE 4.3 A stretched string of finite length L is held fixed at its ends and is subjected

to an initial displacement 0( , 0) sin ( / ).u x u x Lπ= The string is released from this position with

zero initial velocity. Find the resultant time dependent motion of the string.

Solution One of the practical applications of the theory of wave motion is the vibrationof a stretched string, say, that of a musical instrument. In the present problem, let us considera stretched string of finite length L, which is subjected to an initial disturbance. The governingequation of motion is

2PDE: 0, 0 , 0tt xxu c u x L t− = ≤ ≤ > (1)

BCs: (0, ) ( , ) 0u t u L t= = (2)

0ICs: ( , 0) sin ( / ),u x u x Lπ= (3)

( , 0) 0u

xt

=�� (4)

In Section 4.3, we have shown the solution of the one-dimensional wave equation by canonicalreduction as

( , ) ( ) ( )u x t x ct x ct� �� (5)

One of the known methods for solving this problem is based on trial function approach. Letus choose a trial function of the form

( , ) sin ( ) sin ( )u x t A x ct x ctL L

π π⎡ ⎤= + + −⎢ ⎥⎣ ⎦(6)

where A is an arbitrary constant. Now, we rewrite Eq. (6) as

( , ) 2 sin cosx c t

u x t AL L

� ��

(7)

obviously, Eq. (7) satisfies the initial condition (3) with 0 /2,A u= while the second initial

condition (4) is satisfied identically. In fact Eq. (7) also satisfies the boundary condition (2).Therefore, the final solution is found to be

0( , ) sin cosx c t

u x t uL L

� ��

(8)



It may be noted that the trial function approach is easily adoptable if the initial condition isspecified as a sin function. However, it is difficult if the initial conditions are specified as a

general function such as ( ).f x In such case, it is better to follow variables separable method

as explained in Section 4.5.

EXAMPLE 4.4 Solve the following initial value problem of the wave equation (Cauchyproblem), described by the inhomogeneous wave equation

2PDE: ( , )tt xxu c u f x t− =

subject to the initial conditions

( , 0) ( ), ( , 0) ( )tu x x u x v xη= =

Solution To make the task easy, we shall set 1 2 ,u u u� � so that 1u is a solution ofthe homogeneous wave equation subject to the general initial conditions given above. Then

2u will be a solution of

2 222 2

2 2( , )

u uc f x t

t x� �

� ��

(4.20)

subject to the homogeneous ICs

22 ( , 0) 0, ( , 0) 0

uu x x

t= =

��

(4.21)

To obtain the value of u at 0 0( , ),P x t we integrate the partial differential equation (4.20) over

the region IR as shown in Fig. 4.4, to obtain

2 222 2

2 2IR IR

( , )u u

c dx dt f x t dx dtt x

⎛ ⎞− =⎜ ⎟⎜ ⎟⎝ ⎠

∫∫ ∫∫� ��

Using Green’s theorem in a plane to the left-hand side of the above equation to replace the

surface integral over IR by a line integral around the boundary IR of IR,� the above equation

reduces to

2 2 2

IR IR

( , )u u

c dx dt f x t dx dtx x t t

⎡ ⎤⎛ ⎞ ⎛ ⎞− − =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∫∫ ∫∫� ��

and finally to

22 2

IRIR

( , )u u

dx c dt f x t dx dtt x

⎛ ⎞+ =⎜ ⎟⎝ ⎠∫ ∫∫�

� ��

(4.22)



Now, the boundary IR� comprises three segments , and .BP PA AB Along , / ;BP dx dt c= −

along , / .PB dx dt c= Using these results, Eq. (4.22) becomes

2 2 2 2

BP PA

u u u uc dt dx c dt dx

t x t x⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫� � � ��

22 2

IR

( , )AB

u udx c dt f x t dx dt

t x⎛ ⎞− + =⎜ ⎟⎝ ⎠∫ ∫∫� ��

The integrands of the first two integrals are simply the total differentials, while in the thirdintegral, the first term vanishes on AB in view of the second IC in Eq. (4.21), and the second

term also vanishes because AB is directed along the x-axis on which / 0.dt dx = Then we arrive

at the result

2 2

IR

( , )BP PA

c du c du f x t dx dt− =∫ ∫ ∫∫which can be rewritten as

2 2 2 2

IR

( ) ( ) ( ) ( ) ( , )cu P cu B cu P cu A f x t dx dt� � � � �� (4.23)

Using the first IC of Eq. (4.21), we get 2 2( ) ( ) 0,u A u B� � and hence Eq. (4.23) becomes

2

IR

1( ) ( , )

2u P f x t dx dt

c� ��

with the help of Fig. 4.4, we deduce

0 0 0

0 02

0

1( ) ( , )

2

t x ct ct

x ct ctu P f x t dx dt

c

+ −

− += ∫ ∫ (4.24)

Now, using the fact that 1 2 ,u u u� � as also using Eq. (4.24) and D’Alembert’s solution

(4.18), the required solution of the inhomogeneous wave equation subject to the given ICs isgiven by

1 1( , ) { ( ) ( ) ( )

2

x ct


cη η ξ ξ

+

−= + + − + ∫

0 0 0

0 00

1( , )

2

t x ct ct

x ct ctf x t dx dt

+ −

− ++ ∫ ∫ (4.25)

This solution is known as the Riemann-Volterra solution.



�� !�� !"��

Following Tychonov and Samarski, it is known that transverse vibration of a string is normallygenerated in musical instruments. We distinguish the string instruments depending on whetherthe string is plucked as in the case of guitar or struck as in the case of harmonium or piano.In the case of strings which are struck we give a fixed initial velocity but does not undergoany initial displacement. In the case of plucked instruments, the strings vibrate from a fixedinitial displacement without any initial velocity. The vibrations of stretched strings of musicalinstruments, vocal cards, power transmission cables, guy wires for antennae structures, etc.can be examined by considering the basic form of wave equations as discussed in Sections 4.3and 4.4.

Let a thin homogeneous string which is perfectly flexible under uniform tension lie in its

equilibrium position along the x-axis. The ends of the string are fixed at 0x � and .x L� Thestring is pulled aside a short distance and released. If no external forces are present whichcorrespond to the case of free vibrations, the subsequent motion of the string is described by

the solution ( , )u x t of the following problem:

PDE: 2 0,tt xxu c u− = 0 , 0x L t≤ ≤ > (4.26)

BCs: (0, ) 0,u t = 0t �

( , ) 0,u L t = 0t � (4.27)

ICs: ( , 0) ( ),u x f x= ( , 0) ( )tu x g x= (4.28)

To obtain the variables separable solution, we assume

( , ) ( ) ( )u x t X x T t= (4.29)

and substituting into Eq. (4.26), we obtain

2 22

2 2

d T d XX c T

dt dx�

i.e.,

2 2 2 2

2

/ /(a separation constant)

d X dx d T dtk

X c T= =

Case I When 0k � , we have 2. Thenk λ=

22

20

d XX

dx��

22 2

20

d Tc T

dt��



Their solution can be put in the form

1 2x xX c e c e� ��

� � (4.30)

3 4c t c tT c e c e� ��

� � (4.31)

Therefore,

1 2 3 4( , ) ( ) ( )x x c t c tu x t c e c e c e c eλ λ λ λ− −= + + (4.32)

Now, use the BCs:

1 2 3 4(0, ) 0 ( ) ( )c t c tu t c c c e c eλ λ−= = + + (4.33)

which imply that 1 2 0c c� � . Also, ( , ) 0u L t � gives

1 2 0L Lc e c eλ λ− −+ = (4.34)

Equations (4.33) and (4.34) possess a non-trivial solution iff

1 10L L

L Le e

e e� �

� �

�

�

� � �

or

2 21 0 implying 1 or 0L Le e L� ��

This implies that 0,� � since L cannot be zero, which is against the assumption as in CaseI. Hence, this solution is not acceptable.

Case II Let 0.k � Then we have

2 2

2 20, 0

d X d T

dx dt� �

Their solutions are found to be

,X Ax B= + T ct D� �

Therefore, the rquired solution of the PDE (4.26) is

( , ) ( ) ( )u x t Ax B ct D= + +

Using the BCs, we have

(0, ) 0 ( ), implying 0u t B ct D B= = + =

( , ) 0 ( ), implying 0u L t AL ct D A= = + =

Hence, only a trivial solution is possible. Since we are looking for a non-trivial solution,consider the following case.



Case III When 0�k , say 2k �� , the differential equations are

2 22 2 2

2 20, 0

d X d TX c T

dx dt� ��

Their general solutions give

1 2 3 4( , ) ( cos sin ) ( cos sin )u x t c x c x c c t c c tλ λ λ λ= + + (4.35)

Using the BC: (0, ) 0u t = we obtain 1 0c � . Also, using the BC: ( , ) 0,u L t = we get sin 0,L� �

implying that / , 1, 2, ,n n L n� �� which are the eigenvalues. Hence the possible solution is

( , ) sin cos sin , 1, 2,n n nn x n ct n ct

u x t A B nL L L

� � �� (4.36)

Using the superposition principle, we have

1

( , ) sin cos sinn nn

n x n ct n ctu x t A B

L L L

π π π�

�

� �� (4.37)

The initial conditions give

1

( , 0) ( ) sinnn

n xu x f x A

L

π�

��

which is a half-range Fourier sine series, where

0

2( ) sin

L

nn x

A f x dxL L

π= ∫ (4.38)

Also,

1

( , 0) ( ) sint nn

n x nu x g x B c

L L

π π�

�

� �� which is also a half-range sine series, where

0

2( )sin

L

nn x

B g x dxn c L

ππ

= ∫ (4.39)

Hence the required physically meaningful solution is obtained from Eq. (4.37), where nA and

nB are given by Eqs. (4.38) and (4.39). ( , )nu x t given by Eq. (4.36) are called normal modes

of vibration and / , 1, 2,nn c L nπ ω= = … are called normal frequencies.



The following comments may be noted:

(i) The displaced form of the stretched string defined by Eq. (4.36) is referred toas the nth eigenfunction or the nth normal mode of vibration.

(ii) The period of the nth normal mode is 2L/nC, which means nC/2L cycles persecond, called its frequency.

(iii) The frequency can be expressed as

1/2

2

nf

L

tr

Ê ˆ= Á ˜Ë ¯

Thus, the frequency can be increased either by reducing L or by increasing thetension t.

(iv) For a given L, t and r, the first normal mode n = 1, vibrates with the lowestfrequency

2 24

Cf

LL

tr

= = ,

is called the fundamental frequency.

(v) If the stretched string can be made to vibrate in a higher normal mode, thefrequency is increased by an integer multiple. The deflected configuration of thestretched string corresponding to the given normal mode at a specified timet = t*, can be obtained from Eq. (4.36). The deflected shapes corresponding tothe first three normal modes and the associated frequencies are depicted inFig. 4.4.

Fig. 4.4(a) Normal modes of a vibrating stretched string.



EXAMPLE 4.5 Obtain the solution of the wave equation

2tt xxu c u�

under the following conditions:

(i) (0, ) (2, ) 0u t u t= =

(ii) 3( , 0) sin /2u x xπ=

(iii) ( , 0) 0.tu x =

Solution We have noted in Example 4.4 that the physically acceptable solution of thewave equation is given by Eq. (4.35), and is of the form

1 2 3 4( , ) ( cos sin ) [ cos ( ) sin ( )]u x t c x c x c c t c c tλ λ λ λ= + +

Using the condition (0, ) 0,u t = we obtain 1 0.c � Also, condition (iii) implies 4 0.c � The

condition (2, ) 0u t = gives

sin 2 0,λ =

implying that

/2, 1, 2,n nλ π= = …Thus, the possible solution is

1

( , ) sin cos2 2n

n

n x n ctu x t A

π π�

�� (4.40)

Finally, using condition (ii), we obtain

3

1

3 1 3sin sin sin sin

2 2 4 2 4 2nn

n x x x xA

π π π π�

��

which gives 1 33/4, 1/4,A A= = − while all other An’s are zero. Hence, the required solution is

3 1 3 3( , ) sin cos sin cos

4 2 2 4 2 2

x ct x ctu x t

π π π π= −

EXAMPLE 4.6 Prove that the total energy of a string, which is fixed at the points 0,x x L� �

and executing small transverse vibrations, is given by

2 2

20

1 1

2

L y yT dx

x tc

⎡ ⎤⎛ ⎞ ⎛ ⎞+⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦∫ � �

� �



where 2 / ,c T ρ ρ= is the uniform linear density and T is the tension. Show also that if

( ), 0 ,y f x ct x L� � � � then the energy of the wave is equally divided between potential

energy and kinetic energy.

Solution The kinetic energy (KE) of an element dx of the string executing small transversevibrations is given by (see Fig. 4.5)

21( )

2

ydx

tρ ⎛ ⎞

⎜ ⎟⎝ ⎠��

ds

x = 0 x L = Ox

y

Fig. 4.5 Vibrating string.

Therefore,

2

20

1Total KE

2

LT ydx

tc

⎛ ⎞= ⎜ ⎟⎝ ⎠∫ ��

(4.41)

However, 2 2 2 ,ds dx dy� � which gives

2 211 1

2

dy dyds dx dx

dx dx

� ��

Hence, the stretch in the string is given by

21

2

yds dx dx

x⎛ ⎞− = ⎜ ⎟⎝ ⎠��

Now, the potential energy (PE) of this element is given by

21PE

2

yT dx

x⎛ ⎞= ⎜ ⎟⎝ ⎠��

Therefore,

2

0Total PE

2

LT ydx

x⎛ ⎞= ⎜ ⎟⎝ ⎠∫ ��

(4.42)



When added, Eqs. (4.41) and (4.42) will yield the required total energy of the string. If

( ),y f x ct= − then

( ), ( )y ycf x ct f x ctt x

′ ′= − − = −∂ ∂∂ ∂

22 2( )y c f

t⎛ ⎞ ′=⎜ ⎟⎝ ⎠∂∂

From Eq. (4.41),

20

1Total KE ( )2

LT f dx′= ∫

From Eq. (4.42),

20

1Total PE ( )2

LT f dx′= ∫

which clearly demonstrates that the total KE = total PE.

EXAMPLE 4.7 A string of length L is released from rest in the position ( )y f x= . Show

that the total energy of the string is2

2 2

14 sn

T s kL

π ∞

=∑

where

0

2 ( ) sin ( / )L

sk f x s x L dxL

π= ∫T-tension in the string

If the mid-point of a string is pulled aside through a small distance and then released,

show that in the subsequent motion the fundamental mode contributes 28/π of the total

energy.

Solution If ( )f x can be expressed in Fourier series, then

01

( ) ( / 2) ( cos sin )n nn

f x a a nx b nx∞

== + +∑

Here, 1 1( cos sin )a x b x+ is called the fundamental mode. Following the variables separable

method and using the superposition principle, the general solution of the wave equation is

1( , ) sin cosn

n

n x cn ty x t kL Lπ π∞

==∑ (4.43)



where

0

2 ( )sinL

nn xk f x dx

L Lπ= ∫

and the total energy is obtained from

2 2

20

12

LT y yE dxx tc

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

∫∂ ∂∂ ∂

(4.44)

From Eq. (4.43),

1

1

cos cos

sin sin

•

=

•

=

=

=

nn

nn

y n n x cn tkx L L L

y cn n x cn tkt L L L

π π π

π π π−

∑

∑

∂∂

∂∂

Using the standard integrals

2

0

0,sin sin

,m n

mx nx dxm n

π

π≠⎧

= ⎨ =⎩∫

2

0

0,cos cos

,m n

mx nx dxm n

π

π≠⎧

= ⎨ =⎩∫We have

2 22 2 2 2

20 0

22 2 2

cos cos

cos2

L Ln

n

y n x cn tdx k n dxx L LL

cn tk nL L

π π π

π π

⎛ ⎞ =⎜ ⎟⎝ ⎠

=

∑∫ ∫

∑

∂∂

(4.45)

Also,

2 22 2 2

2 20

22 2 2

1 sin2

sin2

Ln

n

y cn t Ldx k nt Lc L

cn tk nL L

π π

π π

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

∑∫

∑

∂∂

(4.46)



Substituting Eqs. (4.45) and (4.46) into Eq. (4.44), we obtain

22 2

12 2 nn

TE n k

L

π �

�� (4.47)

Now, the transverse motion of the string is described by the equation

1

( , ) sin cosnn

n x n cty x t k

L L

π π�

��

where

0

2( ) sin

L

nn x

k f x dxL L

π= ∫But the equation of the line OP is (see Fig. 4.6)

2, 0 /2y x x L

L

ε= ≤ ≤

xO

y

(0, 0) ( , 0)LA

ε

P L( /2, )ε


while the equation for the line of PA is

2( ), /2y x L L x L

L

ε= − − ≤ ≤

Therefore,

/2

0 /2

2 2 2sin ( )sin

L L

nL

n x n xk x dx x L dx

L L L L L

ε π ε π⎡ ⎤= + − −⎢ ⎥⎣ ⎦∫ ∫Integration by parts yields

2 2

2 2 2 2

2 2

2 2 2sin

2

8sin for odd

2

nL L n

kL L Ln n

nn

n

ε ε ππ π

ε ππ

⎡ ⎤= +⎢ ⎥

⎢ ⎥⎣ ⎦

= (4.48)



Substituting this value of nk into Eq. (4.47), we obtain

2 2 22

4 4 2 2odd odd

64 1 16 1

4

� � ��

� �� n n

T TE n

L n L n

but we know that

2

2odd

1

8n n

π=∑Hence,

2 216Total energy,

8

TE

L

ε ππ 2

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

while the total energy due to the fundamental mode is 2 216 / .T Lε π Hence the result.

��# �� "�� !��

Consider the problems of forced vibrations of a finite string due to an external driving force.If we assume that the string is released from rest, from its equilibrium position, the resultingmotion of the string is governed by

PDE: 2 ( , )tt xxu c u F x t− = 0 , 0x L t≤ ≤ ≥ (4.49)

BCs: (0, ) ( , ) 0,u t u L t= = 0t � (4.50)

ICs: ( , 0) ( , 0) 0,tu x u x= = 0 x L� � (4.51)

Here, ( , )F x t is the external driving force. To obtain the solution of the above problem, we

proceed as follows: Taking the solution of vibrating string in the absence of applied externalforces as a guideline, we assume the solution to this case to be

1

( , ) ( )sinnn

n xu x t t

L

πφ�

�� (4.52)

It can be seen easily that the BCs are satisfied. The function ( , )u x t defined by Eq. (4.52)

also satisfies the ICs (4.51), provided.

(0) (0) 0, 1, 2,n n nφ φ′= = = … (4.53)

Substituting the assumed solution (4.52) into the governing PDE (4.49), we obtain

2 22

21

( ) ( ) sin ( , )n nn

n n xt c t F x t

LL

π πφ φ�

�

� ��

� ��



or

2

1

( ) ( ) sin ( , )n n nn

n xt t F x t

L

πφ ω φ�

�

� �� (4.54)

where

nn c

L

�� (4.55)

and the dots over � denote differentiation with respect to t. Multiplying Eq. (4.54) by

sin /k x Lπ and integrating with respect to x from 0x � to x L� and interchanging the order

of summation and integration, we get

2

0 01

[ ( ) ( )] sin sin ( , )sin ( )L L

n n n kn

n x k x k xt t dx F x t dx F t

L L L

π π πφ ω φ�

��

From the orthogonality property of the function sin ( / ),n x Lπ we have

2 2

0[ ( ) ( )] sin ( )

L

k k k kk x

t t dx F tL

πφ ω φ+ =∫��

or

2 2[ ( ) ( )] ( ), 1, 2,k k k kt t F t k

Lφ ω φ+ = = …�� (4.56)

This is a linear second order ODE which, for instance, can be solved by using the methodof variation of parameters. Thus, we solve

2( ) ( ) ( )kk k kt t F t� � �� (4.57)

subject to

(0) (0) 0k kφ φ= =�

where

0

2( ) ( , )sin

L

kk x

F t F x t dxL L

π= ∫The complementary function for the homogeneous part is cos sin .k kA t B t� �� Taking A and

B as functinos of t, let

( ) ( )cos ( )sink k kt A t t B t t� � ��

( ) cos sin sin cosk k k k k k kt A t B t A t B tφ ω ω ω ω ω ω= + − +� � �



We choose A and B such that

cos sin 0k kA t B tω ω+ =� � (4.58)

Therefore,

2 2( ) cos sin sin cosk k k k k k k k kt A t B t A t B tφ ω ω ω ω ω ω ω ω= − − +��

Substituting these expressions into Eq. (4.57), we get

( cos sin ) ( )k k k kB t A t F tω ω ω− =�� (4.59)

Solving Eqs. (4.58) and (4.59) for and ,A B� � we obtain

( )sin( ) k k

k

F t tA t

ωω

= −�

( ) cos( ) k k

k

F t tB t

ωω

=�

Integrating, we get

0

1( ) sin

t

k kk

A F dξ ω ξ ξω

= − ∫

0

1( ) cos

t

k kk

B F dξ ω ξ ξω

= ∫Thus,

0

1( ) sin [ ( )]

t

k kk

F t dφ ξ ω ξ ξω

= −∫ (4.60)

It can be verified easily that zero ICs are also satisfied. Hence the formal solution to theproblem described by Eqs. (4.49) to (4.51), using the superposition principle, is

01

1( , ) ( )sin [ ( )] sin

t

n nnn

n xu x t F t d

L

πξ ω ξ ξω

�

�

� ��

� � � (4.61)

Thus, if 1u is a solution of the problem defined by Eqs. (4.26) to (4.28) and if 2u is a solution

of the problem described by Eqs. (4.49)–(4.51), then 1 2( )u u� is a solution of the IBVP

described by

2PDE: ( , ), 0 , 0

BCs: (0, ) ( , ) 0, 0

ICs: ( , 0) ( ), ( ,0) ( )

tt xx

t

u c u F x t x L t

u t u L t t

u x f x u x g x

− = ≤ ≤ ≥

= = ≥

= =

(4.62)

(4.63)

(4.64)



Hence, the solution of this nonhomogeneous problem is found to be

1

( , ) ( cos sin )sinn n n nn

n xu x t A t B t

L

πω ω�

��

01

1( )sin [ ( )] sin

t

n nnn

n xF t d

L

πξ ω ξ ξω

�

�

� ��

� � � (4.65)

This solution may be termed as formal solution, because it has not been proved that the seriesactually converages and represents a function which satisfies all the conditions of the givenphysical problem.

��$ �� "� ��!��

Let IR be a region in the xy-plane bounded by a simple closed curve IR .� Let IR IR IR .U= �Consider the problem described by

PDE: 2 2 ( , , ), , IR , 0ttu c u F x y t x y t− ∇ = ∈ ≥ (4.66)

BCs: ( ) 0 on IR, 0B u t= ≥� (4.67)

ICs: ( , , 0) ( , ) in IRu x y f x y=

( , , 0) ( , ) in IRtu x y g x y= (4.68)

where ( ) 0B u � stands for any one of the following boundary conditions:

(i) 0 on IRu = � (Dirichlet condition)

(ii) 0 on IRu

n=� �

�(Neumann condition)

(iii) 0 on IRu

un

= =� ��

(Robin/Mixed condition)

Before we discuss the method of eigenfunctions, it is appropriate to introduce Helmholtzequation or the space form of the wave equation. The wave equation in three dimensions maybe written in vectorial form as

2 2ttu c u= ∇

By the variables separable method, we assume the solution in the form

( , , ) ( , , ) ( )u x y z t x y z T tφ=



Substituting into the above wave equation, we obtain

2 2T c T� ��

which gives

2

2(a separation constant)

T

c T

φ λφ

′′ ∇= = −

thereby implying

2 0T c T�� (4.69)

2 0φ λφ∇ + = (4.70)

Equation (4.70) is the space form of the wave equation or Helmholtz equation. Of course,

0� � is the trivial solution Eq. (4.70). But a nontrivial solution � exists only for certain

values of { },n� called eigenvalues and the corresponding solution { },n� are the eigenfunctions.

Corresponding to each eigenvalue ,n� there exists at least one real-valued twice continuously

differentiable function n� such that

2 0 in IR

0 on IR

n n n

n

φ λ φ

φ

∇ + =

= �

It may be noted that the sequence of eigenfunctions ( )n� satisfies the orthogonality property

IR

0 for alln m dA n mφ φ = ≠∫∫As in the case of one-dimensional wave equation, each continuously differentiable function

in IR, which vanishes on IR,� can have a Fourier series expansion relative to the orthogonal

set { }.n� Thus the solution to the proposed problem can be written as

1

( , , ) ( ) ( , )n nn

u x y t C t x yφ�

�� (4.71)

where ( )nC t has to be found out. Substitution of the Fourier series into the PDE (4.66) yields

2 2

1

[ ( ) ( , ) ( ) ( , )] ( , , )n n n nn

C t x y C C t x y F x y tφ φ�

��

But

2n n nφ λ φ∇ = −



Therefore,

2

1

[ ( ) ( )] ( , , )n n n nn

C t C t F x y tω φ�

�� (4.72)

where

2 2 , 1,2, ...n nC n� �� (4.73)

Multiplying both sides of Eq. (4.72) by m� and integrating over the ragion IR and interchaning

the order of integration and summation, we obtain

2

1 IR IR

[ ( ) ( )] ( , ) ( , )n n n n m mn

C t C t x y x y dA F dAω φ φ φ�

��

Using the orthogonality property, this equation can be reduced to

2( ) ( ) ( )m m mC t C t F tω+ =�� (4.74)

where

2IR

1( ) ( , , ) ( , )

|| ||m m

m

F t F x y t x y dAφφ

= ∫∫ (4.75)

and

2 2

IR

|| || | |m m dAφ φ= ∫∫ (4.76)

The series (4.71) satisfies the ICs (4.68) is

(0) ( , ) ( , )n nC x y f x yφ =∑(0) ( , ) ( , )n nC x y g x yφ =∑ �

In order to determine (0) and (0),n nC C� we multiply both sides of the above two equations by

m� and integrate over IR and use the orthogonality property to get

2IR

1(0) ( , ) ( , )

|| ||m m

m

C f x y x y dAφφ

= ∫∫ (4.77)

2IR

1(0) ( , ) ( , )

|| ||m m

m

C g x y x y dAφφ

= ∫∫�



Using the method of variation of parameters, the general solution of Eq. (4.74) is given by

0

1( ) cos sin ( )sin ( )

t

m m m m m m mn

C t A t B t F t dω ω ξ ω ξ ξω

= + + −∫Using Eq. (4.77), we obtain

2IR

1( , ) ( , )

|| ||m m

m

A f x y x y dAφφ

= ∫∫ (4.78)

2IR

1( , ) ( , ) , 1 ,2,

|| ||m m

m m

B g x y x y dA mφω φ

= = …∫∫Hence, the formal series solution of the general problem represented by Eqs. (4.66)–(4.68) is

01 1

1( , , ) ( cos sin ) ( , ) ( ) sin { ( )} ( , )

t

n n n n n n n nnn n

u x y t A t B t x y F t d x yω ω φ ξ ω ξ ξ φω

� �

� �

� �� (4.79)

��% ��

In cylindrical coordinates with u depending only on r, the one-dimensional wave equationassumes the following form:

2

2 2

1 1u ur

r r r c t

⎛ ⎞ =⎜ ⎟⎝ ⎠� � ��

(4.80)

If we are looking for a periodic solution in time, we set

( ) i tu F r e �� (4.81)

Then

22

2( ) , ( )i t i tu u

F r e F r er t

ω ωω′= = −� ��

Inserting these expressions, Eq. (4.80) reduces to

2

2

1[ ( ) ] ( )i t i trF r e F r e

r r cω ωω′ = −�

�



or

2

2

( )( ) ( ) 0

F rF r F r

r c

ω′′′ + + = (4.82)

which has the form of Bessel’s equation and hence its solution can at once be written as

0 0r r

F AJ BYc c

� ��

(4.83)

In complex form, we can write this equation as

1 0 0 2 0 0r r r r

F C J iY C J iYc c c c

� � � �� (4.84)

It can be rewritten as

(1) (2)1 20 0

r rF C H C H

c c

ω ω⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠(4.85)

where (1) (2)0 0,H H are Hankel functions defined by

(1)0 00

r rH J iY

c c

� ��

(4.86)

(2)0 00

r rH J iY

c c

ω ω⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠(4.87)

which behave like damped trigonometric functions for large r. Thus the solution of one-dimensinoal wave equation becomes

(1) (2)1 20 0

i t i tr ru C e H C e H

c c� ��

� � � �(4.88)

Using asymptotic expressions, for (1) (2)0 0andH H defined by

(1) ( /4)0

(2) ( /4)0

2( )

2( ) for large

i x

i x

H x ex

H x e xx

π

π

π

π

−

− −

=

=

(4.89)



the general periodic solution to the given wave equation in cylindrical coordinates is

/4 /41 2

2 exp [ ( / )( ) exp [ ( / ) ( )( , ) i ic i c r ct i c r ct

u r t C e C er r

π πω ωπω

− + −⎡ ⎤= +⎢ ⎥⎣ ⎦(4.90)

��& ��

In spherical polar coordinates, with u depending only on r, the source distance, the waveequation assumes the following form:

22 2 2

1 1,

u ur

r rr c t

2⎛ ⎞ =⎜ ⎟⎝ ⎠� � ��

0r > (4.91)

We look for a periodic solution in time in the form

( ) i tu F r e �� (4.92)

Then

( ) ,i tuF r e

rω′=�

�

22

2( ) i tu

F r et

ωω= −��

Substituting these derivatives into Eq. (4.91), we obtain

22

2 2

1( ) ( )i t i tr F e F r e

rr cω ωω′ = −�

�

i.e.

22

2 2

1[ 2 ]i t i te r F rF e F

r cω ωω′′ ′+ = −

Therefore,

2

2

20F F F

r c

�� (4.93)

Let

1/2

( )F r rc

ω ψ−⎛ ⎞= ⎜ ⎟⎝ ⎠

Then3/2 1/2

2 5/2 3/2 1/2

( ) ( )2

3( ) ( ) ( )

4

F r r r rc c c

F r r r r r rc c c c c

ω ω ωψ ψ

ω ω ω ω ωψ ψ ψ

− −

− − −

⎛ ⎞ ⎛ ⎞′ ′= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞′′ ′ ′′= − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠



Substituting into Eq. (4.93), we obtain

1/2 2 21 1

( ) ( ) ( ) 02

r r r rc r c r

ω ωψ ψ ψ− ⎡ ⎤⎧ ⎫⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞′′ ′⎢ ⎥+ + − =⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭⎣ ⎦

Since 0,rc

��

we have

2 21 1

( ) ( ) ( ) 02

r r rr c r

��

� ��

which is a form of Bessel’s equation, whose solution is given by

1/2 1/2( )r A J r B J rc c

ω ωψ −⎛ ⎞ ⎛ ⎞′ ′= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

where andA B� � are constants. Therefore,

1/2

1/2 1/2( )F r r A J r B J rc c c

ω ω ω−

−⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞′ ′= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

(4.94)

or

1/2 1/2( )A B

F r J r J rr c r c

ω ω−

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠(4.95)

But, we know that

1/22

( ) sinJ x xxπ

=

1/22

( ) cosJ x xxπ− =

Therefore,

2 sin( / ) cos( / )( )

c r c r cF r A B

r r

ω ωπω

⎡ ⎤= +⎢ ⎥⎣ ⎦(4.96)

In complex form,

1 2exp( / ) exp ( / )

( )i r c i r c

F r C Cr r

ω ω−= + (4.97)



Thus, the required solution of the wave equation is

1 2exp ( / )( ) exp ( / )( )

( , )i c r ct i c r ct

u r t C Cr r

ω ω+ − −= + (4.98)

��

To find the solution of the wave equation representing the vibration of a circular membrane,

it is natural that we introduce polar coordinates ( , ), 0 ,0 2 .r r aθ θ π≤ ≤ ≤ ≤ Thus, the governing

two-dimensional wave equation is given by

2 2 2 2 2

1 1 1PDE:

u u u u

r rc t r r

2 2 2

θ= + +� � � �

�� (4.99)

and the boundary and initial conditions are given by

BCs: ( , , ) 0, 0u a t tθ = ≥ (4.100)

i.e. the boundary is held fixed, and

1 2ICs: ( , , 0) ( , ), ( , , 0) ( , )u

u r f r r f rt

θ θ θ θ= =��

(4.101)

Let us look for a solution of Eq. (4.99) in the following variables separable form:

( ) ( ) ( )u R r H T tθ= (4.102)

Substituting into Eq. (4.99), we obtain

2 2

1 1RHTR HT R HT RH T

rc r

��

Dividing throughout by 2/ ,RHT c we get

2 22

1 1(say)

T R R Hc

T R r R Hr�

��

Then

2 0T Tμ′′ + = (4.103)

22 2 2

2(say)

R R Hr r r k

R R Hc

��

i.e.

22 2 2

20

ur R rR r k R

c

� ��

(4.104)



2 0H k H′′ + = (4.105)

Here, 2 2and k� are arbitrary separation constants. The general solutions of Eqs. (4.103)–

(4.105) respectively are

cos sinT A t B t� ��

k kr r

R PJ QYc c

� ��

(4.106)

cos sinH E k F k� ��

where ,k kJ Y are Bessel functions of first and second kind respectively of order k. Thus, the

general solution of the wave equation (4.99) is

( , , ) ( cos sin ) ( cos sin )k kr r

u r t A t B t PJ QY E k F kc c

� ��

� ��

Since the deflection is a single-valued periodic function in � of period 2 , kπ must be integral,

say .k n� Also, since ( / )kY r cμ �� as 0,r� we can avoid infinite deflections at the

centre ( 0)r � by taking 0.Q � Again noting that the BC: (4.100) implies that the deflection u

is zero on the boundary of the circular membrane, we obtain

0na

Jc

��

(4.108)

which has an inifinte number of positive zeros. These zeros (roots) are tabulated for severalvalues of n in many handbooks. Their representation requires two indices. The first oneindicates the order of the Bessel function, and the second, the solution. Thus denoting the

roots by ( 0,1, 2, ; 1, 2, 3, ),nm n mμ = … = … we have, after using the principle of superposition,

the solution of the circular membrane in the form

1 1

( , , ) ( cos sin ) ( cos sin )nmn

m n

ru r t PJ A t B t E n F n

c

μθ μ μ θ θ� �

� �

� �� Alternatively,

� �1 1

( , , ) [ cos sin ]cos [ cos sin ]sin�

� � � � � � ��

� �

� �� nmn nm nm nm nm

m n

ru r t J a n b n t c n d n t

c

(4.109)



Now, to determine the constants, we shall use the prescribed ICs which yield

11 1

( , ) ( cos sin ) nmnm nm n

m n

rf r a n b n J

cμθ θ θ

∞ ∞

= =

⎛ ⎞= + ⎜ ⎟⎝ ⎠∑ ∑ (4.110)

21 1

( , ) ( cos sin ) nmnm nm nm n

m n

rf r c n d n J

cμθ μ θ θ

∞ ∞

= =

⎛ ⎞= + ⎜ ⎟⎝ ⎠∑ ∑Hence, the solution of the circular membrance is given by Eq. (4.109), where

212 2 0 0

212 2 0 0

222 2 0 0

222 2 0 0

2 ( , ) cos[ ( )]

2 ( , ) sin[ ( )]

2 ( , ) cos ,[ ( )]

2 ( , )[ ( )]

anm n nm

n nm

anm n nm

n nm

anm n nm

n nm

anm

n nm

ra f r J n r dr dca J

rb f r J n r dr dca J

rc f r J n r dr dca J

d f ra J

π

π

π

θ μ θ θπ μ

θ μ θ θπ μ

θ μ θ θπ μ

θπ μ

⎛ ⎞= ⎜ ⎟′ ⎝ ⎠

⎛ ⎞= ⎜ ⎟′ ⎝ ⎠

⎛ ⎞= ⎜ ⎟′ ⎝ ⎠

=′

∫ ∫

∫ ∫

∫ ∫

∫ sinn nmrJ n r dr dc

πμ θ θ⎛ ⎞⎜ ⎟⎝ ⎠∫

4.11 UNIQUENESS OF THE SOLUTION FOR THE WAVE EQUATION

In Section 4.5, we have developed the variables separable method to find solutions to thewave equation with certain initial and boundary conditions. A formal solution for the non-homogeneous equation is also given in Section 4.6. In this section, we shall show that thesolution to the wave equation is unique.

Uniqueness Theorem The solution to the wave equation

2 ,tt xxu c u= 0 ,x L< < 0t > (4.111)

satisfying the ICs:

( , 0) ( ), 0

( , 0) ( ), 0t

u x f x x L

u x g x x L

= ≤ ≤

= ≤ ≤

and the BCs:

(0, ) ( , ) 0u t u L t= =

where ( , )u x t is twice continuously differentiable function with respect to x and t, is unique.



Proof Suppose 1 2andu u are two solutions of the given wave equation (4.111) and let

1 2.v u u= − Obviously ( , )v x t is the solution of the following problem:

2 ,tt xxv c v= 0 ,x L< < 0t > (4.112)

( , 0) 0,v x = ( , 0) 0,tv x = 0 x L≤ ≤

and

(0, ) ( , ) 0v t v L t= =

It is required to prove that ( , )v x t is identically zero, implying 1 2.u u� For, let us consider

the function

2 2 2

0

1( ) ( )

2

L

x tE t c v v dx= +∫ (4.113)

which, in fact, represents the total energy of the vibrating string at time .t It may be noted

that ( )E t is differentiable with respect to ,t as ( , )v x t is twice continuously differentiable. Thus,

0

L

x xt t ttdE

c v v v v dxdt

2⎡ ⎤= +⎣ ⎦∫ (4.114)

Integrating by parts, the right-hand side of the above equation gives us

0

2 2 2

0 0

LL L

x xt x t t xxc v v dx c v v c v v dx⎡ ⎤= −⎣ ⎦∫ ∫

But, (0, ) 0v t = implies (0, ) 0tv t = for 0t � and ( , ) 0v L t = implying ( , ) 0tv L t = for 0.t � Hence,

Eq. (4.114) reduces to

2

0( ) 0

Lt tt xx

dEv v c v dx

dt� � ��

In other words, ( ) constant (say).E t c� � Since ( , 0) 0v x = implies ( , 0) 0,xv x = and

( , 0) 0,tv x = we can evaluate c and find that

2 2 2

0 0(0) 0

L

x tt

E c c v v dx=

⎡ ⎤= = + =⎣ ⎦∫

which gives ( ) 0,E t � which is possible if and only if 0xv ≡ and 0tv ≡ for all 0, 0t x L� � �

which is possible only if ( , ) constant .v x t = However, since ( , 0) 0,v x = we find ( , ) 0.v x t ≡

Hence, 1 2( , ) ( , ).u x t u x t= This means that the solution ( , )u x t of the given wave equation

is unique.



��

With the help of Duhamel’s principle, one can find the solution of an inhomogeneous equation,in terms of the general solution of the homogeneous equation. We shall illustrate this principle

for wave equation. Let the Euclidean three-dimensional space be denoted by 3 ,R and a point

in 3R be represented by 1 2 3( , , ).X x x x� If ( , , )v X t τ satisfies for each fixed � the PDE

2 23( , ) ( , ) 0, in ,ttv X t c v X t X R− ∇ =

with the conditions

( , 0, ) 0, ( , 0, ) ( , )tv X v X F Xτ τ τ= =

where ( , )F X � denotes a continuous function defined for X in 3R , and if u satisfies

0( , ) ( , , )

tu X t v X t dτ τ τ= −∫

then ( , )u X t satisfies

2 23( , ), in , 0

( , 0) ( , 0) 0

tt

t

u c u F x t X R t

u X u X

− ∇ = >

= =

Proof Consider the equation

2 2 ( , )ttu c u F X t− ∇ = (4.115)

with

( , 0) ( , 0) 0tu X u X= =

Let us assume the solution of the problem (4.115) in the form

0( , ) ( , , )

tu x t v X t dτ τ τ= −∫ (4.116)

where ( , , )v X t τ τ− is a one-parameter family solution of

2 2 0 for allttv c v τ− ∇ = (4.117)

Further, we assume that at ,t ��

( , 0, ) 0 for all values ofv X τ τ= (4.118)

Now, differentiating with respect to t under integral sign and using the Liebnitz rule, fromEq. (4.116), we have

0( , 0, ) ( , , )

t

t tu v X t v X t dτ τ τ= + −∫



Using Eq. (4.118), we get

0( , , )

t

t tu v X t dτ τ τ= −∫Differentiating this result once again with respect to t, we obtain

0( , 0, ) ( , , )

t

tt t ttu v X t v X t dτ τ τ= + −∫ (4.119)

Noting that u satisfies Eq. (4.115), v satisfies Eq. (4.117), and after using Eq. (4.117), theabove equation reduces to

2 2

0( , 0, )

t

tt tu v X t c v dτ= + ∇∫Finally, using Eq. (4.116), the above equation reduces to

2 2 ( , 0, )tt tu c u v X t− ∇ = (4.120)

Comparing Eqs. (4.115) and (4.120), we obtain

( , 0, ) ( , )tv X t F X t= (4.121)

Therefore, if v satisfies the equation

2 2 0ttv c v− ∇ =

with the conditions

( , 0, ) 0, ( , 0, ) ( , ) attv X v X F X tτ τ τ τ= = =

then, u defined by Eq. (4.116) satisfies the given inhomogeneous equation (4.115) and the

specified conditions. Here, the function ( , )v X t is called the pulse function or the force

function.

EXAMPLE 4.8 Use Duhamel’s principle to solve the heat equation problem described by

( , ) ( , ) ( , ), , 0t xxu x t ku x t f x t x t� � �� (4.122)

( , 0) 0,u x = x��

Solution We have obtained, in Section 3.3, the unique solution of the problem

( , ) ( , ), , 0

( , 0) ( , )

t xxv x t kv x t x t

v x f x τ

� ��

�

in the form

21( , ) exp [ ( ) /(4 )] ( )

4v x t x kt f d

ktξ ξ ξ

π�

��



Hence, using Duhamel’s principle, the solution of the corresponding inhomogeneous problemdescribed by Eq. (4.122) is given by

0( , , ) ( , , )

tu x t v x t dτ τ τ τ= −∫

or

2

0

1 ( )( , , ) exp ( )4 ( ) 4 ( )

t xu x t f d dk t k t

ξτ ξ ξ τπ τ τ

∞

−∞

⎡ ⎤− −= ⎢ ⎥− −⎢ ⎥⎣ ⎦∫ ∫ (4.123)

4.13 MISCELLANEOUS EXAMPLES

EXAMPLE 4.9 A uniform string of line density ρ is stretched to tension 2cρ and executes

a small transverse vibtration in a plane through the undisturbed line of string. The ends

0,x L= of the string are fixed. The string is at rest, with the point x b= drawn aside through

a small distance ε and released at time 0.t = Find an expression for the displacement

( , ).y x t

Solution The transverse vibration of the string is described by

21PDE: xx tty yc

= (4.124)

The boundary and initial conditions are

BCs: (0, ) ( , ) 0y t y L t= =

IC: ( , 0) 0ty x =

Using the variables separable method, let

( , ) ( ) ( )y x t X x T t=

then, we have from Eq. (4.124),

221X T

X Tcλ

′′ ′′= = ±

The equation of the string is given by (see Fig. 4.7)

, 0( ,0)

( ) ,( )

x x bby x x L b x Lb L

ε

ε

⎧ ≤ ≤⎪⎪= ⎨ −⎪ ≤ ≤⎪ −⎩



The solution to the given problem is discussed now for various values of .�

xO

y

(0, 0)( , 0)L

( , 0)bA

ε

P b( , )ε

Fig. 4.7 Illustration of Example 4.9.

Case I Taking the constant 0,� � we have

0X T��


,X Ax B T Ct D� � � �

Therefore,

( , ) ( ) ( )y x t Ax B Ct D� � �

Using the BCs at 0, ,x L� we can observe that 0,A B� � implies a trivial solution.

Case II Taking the constant as 2 ,λ+ we have

2 2 20X X T c T� ��

Thus, the general solution is

( , ) ( cosh sinh ) ( cosh sinh )y x t A x B x C c t D c tλ λ λ λ= + +

Now the BCs:

(0, ) 0 gives 0y t A= =

and

( , ) 0 gives sinh 0y L t B Lλ= =

which is possible only if B = 0. Thus we are again getting only a trivial solution.

Case III If the constant is 2 ,λ� then we have

2 2 20 0X X T c Tλ λ� � � � ��

In this case, the general solution is

( , ) ( cos sin ) ( cos sin )y x t A x B x P c t Q c tλ λ λ λ� � �



using the BCs:

(0, ) 0 gives 0

( , ) 0 gives sin 0

y t A

y L t B Lλ

� �

� �

For a non-trivial solution, 0 .B L n� �� Therefore, / , 1, 2,n L nλ π= = … Also, using the

IC: ( , 0) 0,ty x � we can notice that 0.Q � Hence, the acceptable non-trivial solution is

( , ) sin cos ,n x cn t

y x t BPL L

π π= 1, 2,n = …

Using the principle of superposition, we have

1

( , ) sin cos�

�� n

n

n x cn ty x t b

L L

� ��which gives

1

( , 0) sinnn

n xy x b

L

π�

��

This is half-range sine series, where

0

0 0

2 2 20 0

2 2 2

2( , 0) sin

2 2sin ( ) sin

2 cos ( / ) 2 sin ( )// /

2 cos ( / ) 2 sin( / )( )

( ) / ( ) /

L

n

b L

bb

LL

b b

n xb y x dx

L L

n x n xx dx x L dx

L b L L b L L

n x L n L xx

Lb n L Lb n L

n x L n x Lx L

L b L n L L b L n L

π

ε π ε π

ε π ε ππ π

ε π ε ππ π

=

= + −−

⎡ ⎤⎡ ⎤= − − −⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦

⎡ ⎤⎡ ⎤+ − − − −⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎣ ⎦

∫

∫ ∫

or

2

2 2

2sin

( )n

L n bb

Ln b L b

ε ππ

=−

Hence the subsequent motion of the string is given by

2

2 21

2( , ) sin sin cos

( )n

L n b n x cn ty x t

L L Ln b L b

ε π π ππ

�

��

�� .



EXAMPLE 4.10 Find a particular solution of the problem described by

2PDE: ( ) cos ,tt xxy c y g x tω− = 0 , 0x L t< < >

BCs: (0, ) ( , ) 0,y t y L t= = 0t �

where ( )g x is a piecewise smooth function and � is a positive constant.

Solution Taking the clue from Example 4.9, we assume the solution in the form

1

( , ) ( ) sinnn

n xy x t A t

L

π�

��

To determine ( ),nA t we consider the Fourier sine expansion of ( )g x in the form

1

( ) sinnn

n xg x B

L

π�

��

and substitute into the given PDE which yields

2

1 1

( ) ( ) sin cos sinn n nn n

n c n x n xA t A t t B

L L L

π π πω� �

� �

� ��

Choosing ( )nA t as the solution of the ODE

2

( ) ( ) cosn n nn c

A t A t B tL

��

� ��

we have for any n, the particular solution

( ) cos ifn nn c

A t A tL

��

Therefore,

22

n nn c

A BL

��

� ��

Hence, the required particular solution is given by

2 21

sin( / )( , ) cos

( / )n

n

B n x Ly x t t

n c L

πωπ ω

�

��

�



EXAMPLE 4.11 A rectangular membrane with fastened edges makes free transversevibrations. Explain how a formal series solution can be found.

Solution Mathematically, the problem can be posed as follows:Solve

2PDE: ( ) 0, 0 , 0tt xx yyu c u u x a y b− + = ≤ ≤ ≤ ≤

subject to the BCs:

(i) (0, , ) 0u y t =

(ii) ( , , ) 0u a y t =

(iii) ( , 0, ) 0u x t =

(iv) ( , , ) 0u x b t =

and ICs:

( , , 0) ( , ), ( , , 0) ( , )tu x y f x y u x y g x y� �

We look for a separable solution of the form

( , , ) ( ) ( ) ( )u x y t X x Y y T t=

Substituting into the given PDE, we obtain

22

1(say)

T X Y

T X Yc�

��

Then2 2

2 2

0

(say)

T c T

X Y

X Y

λ

λ μ

′′ + =

′′ ′′+ = − =

thus yielding2 2 20, ( ) 0Y Y X X� � ��

Let 2 2 2 2 2, .p qλ μ μ− = = Then 2 2 2 2.p q r� � � � Therefore, we have

2 2 2 20, 0, 0X p X Y q Y T r c T′′ ′′ ′′+ = + = + =The possible separable solution is

( , , ) ( cos sin ) ( cos sin ) ( cos ( ) sin ( ))u x y t A px B px C qy D qy E rct F rct= + + +

Using the BCs: (0, , ) 0 gives 0

( , 0, ) 0 gives 0

( , , ) 0 gives / , 1, 2,

( , , ) 0 gives / , 1, 2,

u y t A

u x t C

u a y t p m a m

u x b t q n b n

π

π

= =

= =

= = = …

= = = …



Using the principle of superposition, we get

1 1

( , , ) [ cos ( ) sin ( )] sin sinmn mnm n

m x n yu x y t A rct B rct

a b

π π� �

� ��

where

2 22 2 2 2

2 2

m nr p q

a b�

� ��

� �

Applying the initial condition: ( , ,0) ( , ),u x y f x y� Eq. (4.125) gives

1 1

( , ) sin sinmnm n

m x n yf x y A

a b

π π� �

� ��

where

0 0

4( , )sin sin

a b

mnm x n y

A f x y dx dyab a b

π π= ∫ ∫ (4.126)

Finally, applying the initial condition: ( , , 0) ( , ),tu x y g x y� Eq. (4.125) gives

1 1

( , ) sin sinmnm n

m x n yg x y cr B

a b

� ��

� �

� ��where

0 0

4( , ) sin sin

a b

mnm x n y

B g x y dx dyabcr a b

π π= ∫ ∫ (4.127)

Hence, the required series solution is given by Eq. (4.125), where andmn mnA B are given by

Eqs. (4.126) and (4.127).

EXAMPLE 4.12 Solve the IVP described by

2PDE: ( , ), , 0tt xxu c u F x t x t� � ��

with the data

(i) ( , ) 4 ,F x t x t= + (ii) ( , 0) 0,u x = (iii) ( , 0) cosh .tu x bx=

Solution In Example 4.4, we have obtained the Riemann-Volterra solution for theinhomogeneous wave equation in the following form:

IR

1 1 1( , ) [ ( ) ( )] ( ) ( , )

2 2 2

x ct

x ctu x t x ct x ct v d F x t dx dt

c cη η ξ ξ

+

−= − + + + +∫ ∫∫ ��



Since ( , 0) ( ) 0,u x xη= = the first term on the right-hand side of Eq. (4.128) vanishes. Also,

1 1( ) cosh

2 2

1 sin 1[sinh ( ) sinh ( )]

2 2

(cosh sinh ( )) /

x ct x ct

x ct x ct

x ct

x ct

v d b dc c

bb x ct b x ct

c b bc

bx bct bc

ξ ξ ξ ξ

ξ

+ +

− −

+

−

=

⎛ ⎞= = + − −⎜ ⎟⎝ ⎠

=

∫ ∫

and

IR IR

1 1( , ) (4 )

2 2F x t dx dt x t dx dt

c c= +∫∫ ∫∫

From Fig. 4.4, we can write the equation of the line PA in the form

0 0x x ctt

c

� ��

or

0 0x x ct ct� � �

Similarly, the equation of the line PB is

0 0x x ct ct� � �

Thus

0 0 0

0 0

0

0IR

2 2 30 0 0 0 0 0 00

1( , ) (4 )

2

(4 4 ) 2 /6

t x ct ct

x ct ct

t

F x t dt x t dx dtc

x t tx tt t dt x t t

+ −

+ −= +

= − + − = +

∫∫ ∫ ∫

∫The required solution at any point (x, t) is, therefore, given by

32cosh sinh ( )

( , ) 26

bx bct tu x t xt

bc= + +

EXAMPLE 4.13 Derive the wave equation representing the transverse vibration of a stringin the form

22 22

2 21

u u uc

xt x

2 −⎧ ⎫⎪ ⎪⎛ ⎞= +⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

� � ��



Solution Consider the motion of an element PQ s�� of the string as shown in Fig. 4.8.

In equilibrium position, let the string lie along the x-axis, such that PQ is originally at P Q� � .

Let the displacement of PQ from the x-axis, be denoted by u. Let T be the tension in the stringand � be the density of the string. Writing down the equation of motion of the element PQof the string in the u-direction, we have

xO

u

δx

ψ

δs

T

T

Q

Q′

P

P′

Fig. 4.8 An Illustration of Example 4.13.

2

2sin ( ) sin s

uT T

tψ δψ ψ ρδ+ − = �

�

Neglecting squares of small quantities, we get

2

2cos s

uT

tψδψ ρδ= �

�(4.129)

by noting that

22

2tan , sec

u ux

x xψ ψδψ δ= =� �

� �


2 2 23 4

2 2 2cos cos

u u x uT T

st x xρ ψ ψ= =� � � �

�� (4.130)

but

122

2

1cos 1

1 tan

u

xψ

ψ

−⎧ ⎫⎪ ⎪⎛ ⎞= = +⎨ ⎬⎜ ⎟+ ⎝ ⎠⎪ ⎪⎩ ⎭

��

(4.131)



Using Eq. (4.131) into Eq. (4.130), we get

222 2

2 21

u T u u

xt xρ

−⎧ ⎫⎪ ⎪⎛ ⎞= +⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

� � ��

If we define 2 / ,c T ρ= the required wave equation is

222 22

2 21

u u uc

xt x

−⎧ ⎫⎪ ⎪⎛ ⎞= +⎨ ⎬⎜ ⎟⎝ ⎠⎪ ⎪⎩ ⎭

� � ��

(4.132)

This is a non-linear second order partial differential equation.

EXAMPLE 4.14 Using Duhamel’s principle solve the following IBVP:

PDE: ut – uxx = f(x, t), 0 < x < L, 0 < t < BCS: u(0, t) = u(L, t) = 0

IC: u(x, 0) = 0, 0 x L.

Solution Using Duhamel’s principle, the required solution is given by

0( , ) ( , , )

tu x t v x t d� � ��

where v(x, t, �) is the solution of the homogeneous problem described as

vt – vxx = 0, 0 < x < L, 0 < t <

v(0, t, �) = v(L, t, �) = 0

and u(x, 0, �) = f(x, �).

Now, recalling Example 3.17, the solution to this homogeneous problem is obtained as

2( / )

1

( , , ) sinn L tn

n

n xv x t a e

L� ��

��

�

� �� .

Observe that, the Fourier coefficients an depends on the parameter �, so that

0

2( ) ( , ) sin

L

n nn x

a a f x dxL L

�� Hence, the solution to the gives IBVP is found to be

2( / ) ( )

01

( , ) ( ) sint

n L tn

n

n xu x t a e d

L� � ��

��

�

� �� .



��

1. A homogeneous string is stretched and its ends are at 0x � and .x l� Motion isstarted by displacing the string into the form 0( ) sin ( / ),f x u x lπ= from which it is

released at time 0.t � Find the displacement at any point x and time t.2. Solve the boundary value problem described by

2PDE: 0, 0 , 0

BCs: (0, ) ( , ) 0, 0

ICs: ( ,0) 10 sin ( / ), 0

( , 0) 0

tt xx

t

u c u x l t

u t u l t t

u x x l x l

u x

π

− = ≤ ≤ ≥

= = ≥

= ≤ ≤

=

3. Solve the one-dimensional wave equation

2 ,tt xxu c u= 0 , 0x tπ≤ ≤ ≥

subject to

0 when 0 and

0 when 0 and ( , 0) , 0t

u x x

u t u x x x

π

π

� � �

� � � � �

4. Solve2 ,tt xxu c u� 0 , 0x l t� � �

subject to

(0, ) 0,u t = ( , ) 0 for allu l t t=

( , 0) 0,u x = 3( , 0) sin ( / )tu x b x lπ=5. Solve the vibrating string problem described by

2PDE: 0,tt xxu c u− = 0 , 0x l t< < >

BCs: (0, ) ( , ) 0,u t u l t= = 0t �

ICs: ( , 0) ( ),u x f x= 0 x l� �

( , 0) 0,tu x = 0 x l≤ ≤

6. In spherical coordinates, if u is a spherical wave, i.e. ( , ),u u r t� then the waveequation becomes

22

2 2 2

1 1u ur

r rr c t

⎛ ⎞ =⎜ ⎟⎝ ⎠� � ��

which is called the Euler-Poisson-Darboux equation. Find its general solution.



7. Solve the initial value problem described by

2PDE: xtt xxu c u e� �

with the given data

2( ,0) 5, ( ,0)tu x u x x� �

8. Sovle the initial value problem described by

2PDE: ttt xxu c u xe− =

with the data

( , 0) sin , ( , 0) 0tu x x u x= =

9. Solve the initial boundary value problem described by

2PDE: , 0, 0tt xxu c u x t= > >

with the data:

( , 0) 0,u x = ( , 0) 0, 0tu x x� �

(0, ) sin ,u t t= 0t �

10. Determine the solution of the one-dimensional wave equation

2 2

2 2 2

10, 0 , 0x a t

x c t

φ φ− = < < >� ��

with c as a constant, under the following initial and boundary conditions:

(i)/ , 0

( , 0) ( )( ) /( ),

x b x bx f x

a x a b b x aφ

≤ ≤⎧= = ⎨ − − < ≤⎩

(ii) ( , 0) 0, 0x x at

φ = < <��

(iii) (0, ) ( , ) 0, 0.t a t tφ φ= = ≥

11. A piano string of length L is fixed at both ends. The string has a linear density �and is under tension .� At time t = 0, the string is pulled a distance s from equilibriumposition at its mid-point so that it forms an isosceles triangle and is then released

( ).s L� Find the subsequent motion of the string.

12. Obtain the normal frequencies and normal modes for the vibrating string of Problem 11.



13. A flexible stretched string is constrained to move with zero slope at one end 0,x = while

the other end x L= is held fixed against any movement. Find an expression for thetime-dependent motion of the string if it is subjected to the initial displacement givenby

0( , 0) cos2

xy x yL

π⎛ ⎞= ⎜ ⎟⎝ ⎠

and is released from this position with zero velocity.14. Show that if f and g are arbitrary functions, then

( ) ( )u f x vt i y g x vt i yα α= − + + − −

is a solution of the equation

21

xx yy ttu u uc

+ =

provided 2 2 21 / .v cα = −Choose the correct answer in the following questions (15 and 16):

15. The solution of the initial value problem

4 ,tt xxu u= 0,t > x−∞ < < ∞

satisfying the conditions ( , 0) ,u x x= ( , 0) 0 istu x =

(A) x (B) x2/2 (C) 2x (D) 2t (GATE-Maths, 2001)

16. Let ( , )u x tψ= be the solution to the initial value problem

tt xxu u= for , 0x t−∞ < < ∞ >

with ( , 0) sin , ( , 0) cos ,tu x x u x x= = then the value of ( /2, /6)ψ π π is

(A) 3/2 (B) 1/2 (C) 1/ 2 (D) 1 (GATE-Maths, 2003)

17. Solve the following IBVP

PDE: ut = uxx + f(x, t), 0 < x < p, 0 < t < •

BCS: u(0, t) = u(p, t) = 0

IC: u(x, 0) = 0, 0 £ x £ p

Using Duhamel’s principle.


��

Consider the differential equation

( ) ( )Lu x f x� (5.1)

where L is an ordinary linear differential operator, ( )f x is a known function, while ( )u x is

an unknown function. To solve the above differential equation, one method is to find the

operator 1L� in the form of an integral operator with a kernel ( , )G x ξ such that

1( ) ( ) ( , ) ( )u x L f x G x f dξ ξ ξ−= = ∫ (5.2)

The kernel of this integral operator is called Green’s function for the differential operator.Thus the solution to the non-homogeneous differential equation (5.1) can be written down,once the Green’s function for the problem is known. Applying the differential operator L toboth sides of Eq. (5.2), we get

1( ) ( ) ( , ) ( )f x LL f x LG x f dξ ξ ξ−= = ∫ (5.3)

This equation is satisfied if we choose ( , )G x � such that (see propety III of Section 3.4)

( , ) ( )LG x x� � �� (5.4)

where ( )x� �� is a Dirac -function.δ The solution of Eq. (5.4) is called a singularity

solution of Eq. (5.1). In Section 3.4, we have already introduced the concept of Dirac -functionδand studied its various properties. Now, they become handy to understand more about Green’sfunction.

282

��

Green’s Function


GREEN’S FUNCTION 283

Definition 5.1 Let us consider an auxiliary function ( )x� of a real variable x which possesses

derivatives of all orders and vanishes outside a finite interval. Such functions are called testfunctions.

Now we shall introduce the concept of the derivative of a -functionδ in terms of the

derivative of a test function. We say that ( )x� � is the derivative of ( )x� if

( ) ( ) (0)x x dxδ φ φ�

�� (5.5)

for every test function ( ).x� Similarly, we define ( )x� �� by

( ) ( ) (0)� � ��

�� x x dx (5.6)

With this definition of a derivative, we can show that the -functionδ is the derivative of

a Heaviside unit step function ( )H x defined by

1 for 0( )

0 for 0

xH x

x

��

��(5.7)

To see this result, we consider

0( ) ( ) ( ) ( ) ( ) (0)H x x dx H x x dx x dxφ φ φ φ

� � �

��

By comparing the above result with property III of Section 3.4, i.e., with

( ) ( ) (0)x x dxδ φ φ�

��

we obtain

( ) ( )H x x�� (5.8)

Similarly, the notion of -functionδ and its derivative enables us to give a meaning to the

derivative of a function that has a Jump discontinuity at x �� of magnitude unity. Let

1,( )

0,

xH x

x

��

�

��

��

Then, for any test function ( ),xφ we have

( ) ( ) ( ) ( ) ( )H x x dx x x dxξ φ δ ξ φ φ ξ� �

�� (5.9)



It can also be noted that

[( ) ( )] ( ) ( ) ( )d

x H x x H x H xdx

� � � � ��

Integrating both sides with respect to x between the limits to ,�� we get

( ) ( ) ( ) ( ) ( )

( )

x H x x H x dx H x dx

H x dx

ξ ξ ξ ξ ξ

ξ

� �

��

�

��

� � � � � � ��

� �

� �

� (5.10)

We now consider an example to illustrate the inversion of a differential operator byconsidering the BVP:

2

2( ), (0) (1) 0

d uf x u u

dx� � �

In this case, Eq. (5.4) becomes

2

2( )

d GLG x

dxδ ξ= = − (5.11)

Noting that the -functionδ is the derivative of the Heaviside unit step function and integrating

Eq. (5.11), we get

1[ ( , )] ( ) ( )d

G x H x Cdx

� � ��

where 1 ( )C � is an arbitrary function. Integrating the above result once again with respect to

x, we get

1 2

1 2

( , ) ( ) ( ) ( )

( ) ( ) ( ) ( )

G x H x dx C x C

x H x C x C

ξ ξ ξ ξ

ξ ξ ξ ξ

= − + +

= − − + +

∫

where 1 2( ) and ( )C C� � can be determined from the boundary conditions. Thus, from Eq. (5.2) we

have

1 20

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )x

u x x H x f d x C f d C f dξ ξ ξ ξ ξ ξ ξ ξ ξ ξ� �

��



Now, using the boundary condition: (0) 0,u � we get

20 0 0 ( ) ( )C f dξ ξ ξ�

��

implying that 2 ( ) 0.C � � Using the second boundary condition: (1) 0,u � we have

1

10

0 (1 ) ( ) ( ) ( )f d C f dξ ξ ξ ξ ξ ξ�

��

implying that 1( ) (1 ), 0 1,C ξ ξ ξ= − − ≤ ≤ and zero for all other values of .� Hence,

1

0 0( ) ( ) ( ) ( ) (1 ) ( )

xu x x H x f d x f dξ ξ ξ ξ ξ ξ ξ= − − − −∫ ∫ (5.12)

Comparing this result with Eq. (5.2), we have the kernel of the integral operator, which isknown as Green’s function or source function given by

( , ) ( ) ( ) (1 ), 0 1G x x H x xξ ξ ξ ξ ξ= − − − − ≤ ≤ (5.13)

satisfying the boundary conditions: (0, ) (1, ) 0.G Gξ ξ= =This concept can be extended to partial differential equations also. To make the ideas clearer,let us consider

[ ( )] ( )L u f�X X (5.14)

where L is some linear partial differential operator in three independent variables x, y, z andX is a vector in three-dimensional space. Then the Green’s function may be denoted

by ( ; )G ′X X which satisfies the equation

[ ( ; )] ( )L G δ′ ′= −X X X X (5.15)

On expansion, this equation becomes

[ ( , , ; , , )] ( ) ( ) ( )L G x y z x y z x x y y z zδ δ δ′ ′ ′ ′ ′ ′= − − − (5.16)

Here the expression ( )δ ′−X X is the generalization of the concept of delta function in three-

dimensional space IR and ( ; )G �X X represents the effect at the point X due to a source

function or delta function input applied at �X . Equation (5.15) has the following interpretation

in heat conduction or electrostatics: ( ; )G �X X can be viewed as the temperature (the electrostatc

potential) at any point X in IR due to a unit source (due to a unit charge) located at .�XMultiplying Eq. (5.15) on both sides by ( )f �X and integrating over the volume V with

respect to ,�X we get

( ; ) ( ) ( ) ( ) ( )X X

X XV V

L G f dV f dV fδ′ ′

′ ′⎡ ⎤′ ′ ′ ′= − =⎢ ⎥⎣ ⎦∫ ∫X X X X X X X



Comparing with Eq. (5.14), we arrive at

( ) ( ; ) ( )X

XV

u G f dV′

′′ ′= ∫X X X X (5.17)

which is the solution of Eq. (5.14). This leads to the simple definition that a function

( , , ; , , )u x y z x y z′ ′ ′ is a fundamental solution of the equation, for example, 2 0,u∇ = if u is a

solution of the non-homogeneous equation

2 ( , , ; , , )u x y z x y zδ ′ ′ ′∇ =

This idea can be easily extended to higher dimensions. Thus the Green’s function techniquecan be applied, in principle, to find the solution of any linear non-homogeneous partialdifferential equation. Although a neat formula (5.17) has been given for solution of non-homogeneous PDE, in practice, it is not an easy task to construct the Green’s function. Weshall now present a few singularity solutions, also called fundamental solutions to the well-known operators. These will guide us to construct Green’s function for the solutions of partialdifferential equations which occur most frequently in mathematical physics.

To start with, the fundamental solution for a three-dimensional potential problem satisfies

2 ( )u δ∇ = X (5.18)

or

div grad ( )u δ= X

where u can be interpreted, for example, as the electrostatic potential. We seek a solution

which depends only on the source distance | |;r � X thus, for 0, ( )r u r> satisfies

2 22

10

uu r

r rr

⎛ ⎞∇ = =⎜ ⎟⎝ ⎠� ��


Au B

r� �

Using the fact that the potential vanishes at infinity, we have

u A/r�

Integrating Eq. (5.18) over a small sphere R�

of radius ,� we have

[div grad ] 1R

u dVε

=∫Using the divergence theorem, we obtain

1r

udS

rεσ ε==∫ �

�



where �

� is the surface of .R�

Hence,

22 2

14 4 1

AA dS A

r επε π

ε− = − × = − =∫

Therefore,1/4A π= −

Thus, the singularity solution or the fundamental solution of 2 0u∇ = is

1

4u

r��

(5.19)

The two-dimensional case of Eq. (5.18) is

10, 0

ur r

r r r⎛ ⎞ = >⎜ ⎟⎝ ⎠

� ��


lnu A r B� �

Integrating Eq. (5.18) over a disc R�

of small radius ,� whose surface is ,�

� we get

div grad 1R

uε

=∫Hence

2 1r

u A AdS dS

r rεσ εεπε

ε== = × =∫ ∫�

�

Therefore, 1 2 .A / �� The constant B remains arbitrary and can be set equal to zero for

convenience. Thus, the fundamental solution is

1( ) ln

2u r r

�

� (5.20)

If ( , , ) and ( , , )r x y z r x y z′ ′ ′ ′ are two distinct points in three-dimensional space IR, then the

singularity solution of Laplace equation is

1

4 | |u

π=

′−r r(5.21)

Similarly, the singularity solution for the diffusion equation

2 0tu k u− ∇ =

in three-space variables assumes the form

2

3/2

1 | |exp

4 ( )8[ ( )] k tk t τπ τ

⎡ ⎤′− −⎢ ⎥

−− ⎣ ⎦

r r(5.22)



For Helmholtz equation in three space variables, viz.

2 2 0u k u∇ + =the singularity solution is

| |

| |

ike ��

��

r r

r r(5.23)

Before we attempt to solve Eq. (5.17), we shall examine the form of three-dimensional

-functionδ in general curvilinear coordinates as a preparation to study the solution of partial

differential equations in polar coordinates, spherical polar coordinates etc. Suppose we are

looking for a transformation from Cartesian coordinates, x, y to curvilinear coordinates ,ξ ηthrough the relations

( , ), ( , )x f y g� � � �� (5.24)

where f and g are single-valued, continuously differentiable functions of their arguments.

Suppose that under this transformation, 1 2andξ β η β= = correspond to 1 2andx yα α= =respectively. Also,

( , )

( , )

f fdx d df g dJ

dg gdy d d

��

��

� ��

� ��

��

If we transform the coordinates following Eq. (5.24), the relation

1 2 1 2( , ) ( ) ( ) ( , )x y x y dx dyφ δ α δ α φ α α− − =∫∫becomes

1 2 1 2( , ) [ ( , ) ] [ ( , ) ] | | ( , )f g f g J d dφ δ ξ η α δ ξ η α ξ η φ α α− − =∫∫ (5.25)

where J is the Jacobian of the transformation. Equation (5.25) states that the Dirac -functionδ

1 2[ ( , ) ] [ ( , ) ] | |f g J� � � � � � � �� assigns to any test function ( , )f g� the value of that test

function at the points where 1 2, ,f g� �� i.e. at the poins 1 2, .� � � �� Thus we may write

1 2 1 2 1 2[ ( , ) ] [ ( , ) ] | | ( ) ( ) | | ( ) ( )f g J x y Jδ ξ η α δ ξ η α δ α δ α δ ξ β δ η β− − = − − = − −

Hence,

1 21 2

( ) ( )( ) ( )

| |x y

J

δ ξ β δ η βδ α δ α − −− − = (5.26)

In the next few sections, we shall duscuss the Green’s function method for solving partialdifferential equations with particular emphasis on elliptic equations. The discussion on waveequation and heat equation is also included in Sections 5.5 and 5.6 respectively.



5.2 GREEN’S FUNCTION FOR LAPLACE EQUATION

To find analytical solution of the boundary value problems, the Green’s function method isone of the convenient techniques. In this section, we shall give the definition of Green’sfunction for the Laplace equation and study its basic properties. To begin with, we shall

define Green’s function for the Dirichlet problem, i.e., we shall find u such that 2 0u∇ = is

valid inside some finitely bounded region IR, enclosed by IR,∂ a sufficiently smooth surface,

when u f= is prescribed on the boundary IR∂ (see Fig. 5.1).

Fig. 5.1 Region and boundary surface.

Suppose that u is known at every point of the boundary IR∂ and that it satisfies the relation

2 0 in IRu∇ =

The task is to find ( )u P when IR.P ∈ Let OP = r, and let C be a sphere with centre at P

and radius .ε Also, let Σ be the new region exterior to C and interior to IR. Further, let the

boundary of Σ be denoted by Σ,∂ and

1| |

u′ =′−r r (5.27)

where ′r is another point Q either in Σ or on the boundary .Σ∂ If u and u¢ are twice continuously

differentiable functions in IR and have first order derivatives on IR,∂ then by Green’s theorem in

the region IR we have, from Eq. (2.19), the relation

2 2

IR IR

( ) u uu u u u dV u u dSn n′⎛ ⎞′ ′ ′∇ − ∇ = −⎜ ⎟

⎝ ⎠∫∫∫ ∫∫∂

∂ ∂∂ ∂

Here, n is the unit vector normal to dS drawn outwards from IR and n/∂ ∂ denotes differentiation

in that direction. Since 2 2 0u u′∇ = ∇ = within ,Σ∂ we have, in the region ,Σ the relation

IR

0C

u u u uu u dS u u dSn n n n′ ′⎛ ⎞ ⎛ ⎞′ ′ ′− + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ ∫∫∂

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

(5.28)



Inserting u� from relation (5.27), the above equation reduces to

IR

1 1( ) ( )

| | | |

1 1( ) ( ) 0

| | | |

C

uu dS

n n

uu dS

n n

⎡ ⎤⎛ ⎞′ ′ ′−⎢ ⎥⎜ ⎟′ ′− −⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞′ ′+ − =⎢ ⎥⎜ ⎟′ ′− −⎝ ⎠⎣ ⎦

∫∫

∫∫

r rr r r r

r rr r r r

�

� ��

� ��

(5.29)

When Q is on C, we have

2

1 1 1 1,

| | | |nε ε⎛ ⎞= =⎜ ⎟′ ′− −⎝ ⎠r r r r

��

Also, ,dS � the surface element on C, is 2 sin . And on ,d d Cε θ θ φ

( ) ( )u u du� � �r r

or

( ) ( ) ( ) sin cos sin sin cosu u u u u u

u u x y z ux y z x y z

ε θ φ θ φ θ⎛ ⎞ ⎛ ⎞′ = + + + = + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r r r� � � � � ��

Therefore,

( ) ( ) ( ) on

( ) ( ) ( )

u u O C

u uO

n n

ε

ε

′ = +

′ = +

r r

r r� ��

Now,

22

2

0 0

1 1( ) [ ( ) ( )] sin

| |

( ) sin ( )

( ) sin ( )

4 ( ) ( )

C C

C

u dS u O d dn

u P d d O

u P d d O

u O

π π

φ θ

ε ε θ θ φε

θ θ φ ε

θ θ φ ε

π ε

= =

⎛ ⎞′ ′ = + ×⎜ ⎟′−⎝ ⎠

= +

= +

= +

∫∫ ∫∫

∫∫

∫ ∫

r rr r

r

��

21 1( ) ( ) 0 ( ) sin 0 ( )

| |C

udS d d

n nε ε θ θ φ ε

ε⎡ ⎤′ ′ = + +⎢ ⎥′− ⎣ ⎦∫∫ ∫∫r r

r r� ��



Employing these results and taking the limit as 0,� � Eq. (5.29) becomes

IR

1 14 ( ) ( ) ( ) 0

| | | |u u u dS

n nπ ⎡ ⎤⎛ ⎞′ ′+ − =⎢ ⎥⎜ ⎟′ ′− −⎝ ⎠⎣ ⎦∫∫r r r

r r r r�

� ��

implying thereby

IR

1 1 1( ) ( ) (

4 | | | |

uu u dS

n nπ⎡ ⎤⎛ ⎞′ ′= −⎢ ⎥⎜ ⎟′ ′− −⎝ ⎠⎣ ⎦∫∫r r r )

r r r r�

� ��

(5.30)

Therefore, the value of u at an interior point of the region IR is determined, if u and u n/� � are

known on the boundary, IR .� This leads to the conclusion that both the values of u and u n/� � are

required to obtain the solution of Dirichlet’s problem. But this is not so, as can be seen from

the concept of the Green’s function defined as follows: Let ( , )H �r r be a function harmonic

in IR. Then the Green’s function for the Dirichlet problem involving the Laplace operator isdefined by G, the two point function of position, as

1( , ) ( ,

| |G H′ ′= +

′−r r r r )

r r(5.31)

where ( ,H �r r ) satifies the following

(i)

2 2 2

2 2 2( , ) 0H

x y z

⎛ ⎞′+ + =⎜ ⎟⎜ ⎟′ ′ ′⎝ ⎠

r r� � ��

(5.32)

(ii)

1( , ) 0 on IR

| |G H ′= + =

′−r r

r r� (5.33)

Thus the Green’s function for the Dirichlet problem involving the Laplace operator is a

function ( , )G �r r which satisfies the following properties:

(i) 2 ( , ) ( ) in IRG δ′ ′∇ = −r r r r (5.34)

(ii) ( , ) 0 on IRG ′ =r r � (5.35)

(iii) G is symmetric, i.e.,

( , ) ( , )G G′ ′=r r r r (5.36)

(iv) G is continuous, but G n/� � has a discontinuity at the point r, which is given by theequation

0Lt 1

C

GdS

nε →=∫∫ �

� (5.37)



Following the procedure adopted in the derivation of Eq. (5.30), replacing u� by ( , ),G �r rwe can show that

IR

1( ) ( , ) ( ) ( ) ( , )

4

u Gu G u dS

n nπ⎡ ⎤′ ′ ′ ′= − −⎢ ⎥⎣ ⎦∫∫r r r r r r r

�

� ��

(5.38)

From Eqs. (5.31) and (5.33) we can see that 0 on IR.G = � Thus the solution u at an interior

point is given by the relation

IR

1( ) ( ) ( , )

4

Gu u dS

nπ′ ′= − ∫∫r r r r

�

��

(5.39)

and, therefore, the solution of the interior Dirichlet’s problem is reduced to the determination

of Green’s function ( , ).G �r r

Green’s function can be interpreted physically as follows: Let IR� be a grounded electricalconductor (boundary potential zero) and if a unit charge is located at the source point r, then

G is the sum of the potential at the point �r due to the charge at the source point r in free

space and the potential due to the charges induced on IR.� Thus,

1( , ) ( , )

| |G H� ��

��r r r r

r r (5.40)

Hence, property (i), viz, Eq. (5.34), essentially means that 2 0G∇ = everywhere except at thesource point (r).

EXAMPLE 5.1 Consider a sphere with centre at the origin and radius ‘a’. Apply the divergencetheorem to the sphere and show that

2 14 ( )r

rπδ⎛ ⎞∇ = −⎜ ⎟⎝ ⎠

where ( )r� is a Dirac delta function.

Solution Applying the divergence theorem to

1 1grad

r r⎛ ⎞ ⎛ ⎞= ∇⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

we get

1 1ˆ

V S

dV n dSr r

⎛ ⎞ ⎛ ⎞∇ ⋅∇ = ∇ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫∫∫ ∫∫



where n̂ is an outward drawn normal. If ( , , ),u u r � �� then

1 1ˆ ˆ ˆgrad sinr

u u uu e e e

r r rθ φ θ

θ φ+ +�

� � ��

�

Hence,

22 2

1 1 1 1ˆgrad 4 4r

S S S

e dS dS dS ar r r r a

π π⎛ ⎞ ⎛ ⎞⋅ = = = − × = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫∫ ∫∫ ∫∫��

Thus, we observe that 2 (1/ )r∇ has the following properties:

(i) It is undefined at the origin.

(ii) It vanishes if 0.r �

(iii) Its integral over any sphere with centre at the origin is 4 .�� Hence, we concludethat

2 14 ( )

V

dV rr

πδ⎛ ⎞∇ = −⎜ ⎟⎝ ⎠∫∫∫Now we shall prove the symmetric property through the following theorem.

Theorem 5.1 Show that the Green’s function ( , )G �r r has the symmetric property. In other

words, if 1 2andP P are two points within a fininte region IR bounded by the surface IR,� then

the value at 2P of the Green’s function for the point 1P and the surface IR� is equal to the

value at 1P of the Green’s function for the point 2P and the surface IR.�

Proof We have seen in Example 5.1 that 2 (1/ ) 4 ( ).r rπδ∇ = − If we define

1

| |G H� �

��r r

where H is harmonic, then

2 2 214 ( ) 0

| |G H πδ⎛ ⎞ ′∇ = ∇ + ∇ = − − +⎜ ⎟′−⎝ ⎠

r rr r

Recalling Green’s theorem

2 2

IR IR

( )u u

u u u u dV u u dSn n

′⎛ ⎞′ ′ ′∇ − ∇ = −⎜ ⎟⎝ ⎠∫∫∫ ∫∫�

� ��

(5.41)



and taking 1( , ) (1/| |)u G H′ ′= = − +r r r r (where �r is a position vector of Q, a variable point,

and G is a Green’s function for the point 1P (see Fig. 5.2), we have

Fig. 5.2 An illustration of Theorem 5.1.

1( , ) 0 on IR when IRG ′ ′= ∈r r r�

Also,

2 21 1

1

1( , ) 4 ( )

| |G πδ⎛ ⎞′ ′∇ = ∇ = − −⎜ ⎟′−⎝ ⎠

r r r rr r

Similarly, taking 2( , ),u G� �� r r such that IR,��r we get

2

22 2

( , ) 0 on IR

( , ) 4 ( , )

G

G πδ

′ =

′ ′∇ = −

r r

r r r r

�

Substituting these results into Eq. (5.41), we obtain

2 21 2 2 1 1 2

IR IR

2 1

[ ( , ) ( , ) ( , ) ( , )] ( , ) ( , )

( , ) ( , )

GG G G G dV G

n

GG dS

n

⎡′ ′ ′ ′ ′ ′∇ − ∇ = ⎢⎣

⎤′ ′− ⎥⎦

∫∫∫ ∫∫r r r r r r r r r r r r

r r r r

�

��

��

implying thereby

1 2 2 1

IR

4 [ ( , ) ( ) ( , ) ( )] 0G G dVπ δ δ′ ′ ′ ′− − − − =∫∫∫ r r r r r r r r

Using the property of Dirac -function,δ we get immediately the relation

1 2 2 1( , ) ( , )G G=r r r r (5.42)

Property (iv) of the Green’s function is proved in the following theorem.



Theorem 5.2 If G is continuous and G n/∂ ∂ has discontinuity at r, in particular we canshow that

0Lt 1

C

G dSnε →

=∫∫∂

∂∂

Proof Let C be a sphere with radius ε bounded by C∂ (see Fig. 5.1). We have alreadynoted that G satisfies

2 ( )G δ ′∇ = −r r

Integrating both sides over the sphere C, we get

2 1C

G dV∇ =∫∫∫which can also be written as

20

Lt 1C

G dVε →

∇ =∫∫∫Applying the divergence theorem, we get at once the result

0Lt 1

C

G dSnε →

=∫∫∂

∂∂ (5.43)

Now, we shall present two well-known methods:

1. The method of images in Section 5.3.2. The eigenfunction method in Section 5.4 for constructing Green’s function for boundary

value problems.

5.3 THE METHODS OF IMAGES

The method of images has been extensively used in the development of Electrostatics. Itrequires that we examine the effect of a certain source type of singularity at some point P of

a given region IR, together with the influence of another source type of singularity located

at a point P′ outside the region of interest. Here, P′ is the optical image of P in the boundaryof the given region. This approach will work only for very simple geometries. We shallillustrate its application through the following examples for the construction of Green’s function.

EXAMPLE 5.2 Use Green’s function technique to solve the Dirichlet’s problem for a semi-infinite space.

Solution Let the semi-infinite space be defined by 0,x ≥ i.e., 0 , ,x y≤ ≤ ∞ −∞ ≤ ≤ ∞

.z−∞ ≤ ≤ ∞ We have to find a function u such that 2 0 on 0, and ( , )u x u u y z∇ = ≥ = on

0;x = also as .0 ru → ∞→



The Green’s function ( , )G ′r r for the present problem must satisfy the following relations:

(i)1

( , )| |

G H′ = +′−

r rr r

(ii)2 2 2

2 2 2( , ) 0H

x y z

⎛ ⎞′+ + =⎜ ⎟⎜ ⎟′ ′ ′⎝ ⎠

r r� � ��

(iii) ( , ) 0G � �r r on the plane 0.x �

Let ( )P� � be the image of the point ( )P r in 0.x � If condition (ii) is satisfied

1( , )

| |H � � �

��

r rr�

Since PQ P Q�� whenever Q lies on 0,x � we get (see Fig. 5.3)

| | | |′ ′− = −r r r�


Then the required Green’s function by the method of images is given by the equation

1 1( , )

| | | |G ′ = −

′ ′− −r r

r r r�(5.44)

which satisfies condition (iii). But from Eq. (5.39), we have

IR

1( ) ( ) ( , )

4

Gu u dS

nπ′ ′= − ∫∫r r r r

�

��

(5.45)



where dS is the surface element of IR.� Also,

2 2 2 2 2 2

2 2 2 3/20

1 1

( ) ( ) ( ) ( ) ( ) ( )

2

[ ( ) ( ) ]x

G

n x x x y y z z x x y y z z

G x

x x y y z z′=

⎡ ⎤= − −⎢ ⎥′ ′ ′ ′ ′ ′ ′− + − + − + + − + −⎣ ⎦

⎛ ⎞ =⎜ ⎟′⎝ ⎠ ′ ′+ − + −

� ��

��

Substituting this result and noting that ( ) ( , ),u f y z′ ′ ′=r Eq. (5.45) reduces to

2 2 2 3/2

( , )( , , )

2 [ ( ) ( ) ]

x f y z dy dzu x y z

x y y z zπ� �

��

� � � �� (5.46)

which can be integrated when the nature of the function ( , )f y z� � is explicity given.

EXAMPLE 5.3 Obtain the solution of the interior Dirichlet problem for a sphere using theGreen’s function method and hence derive the Poisson integral formula.

Solution The task is to determine the function ( , , )u r � � satisfying

2 0, 0 , 0 , 0 2u r a θ π φ π∇ = ≤ ≤ ≤ ≤ ≤ ≤ (5.47)

subject to

( , , ) ( , )u a fθ φ θ φ= (5.48)

Green’s function for a sphere can be expressed as

1( , ) ( , )

| |G H� ��

��r r r r

r r (5.49)

where H is so chosen that the conditions

2 2 2

2 2 2( , ) 0H

x y zr r

� ��

� � ��

(5.50)

( , ) 0G � �r r (5.51)

are satisfied on the surface of the sphere.

Let ( , , )P r θ φ be a point inside a sphere as shown in Fig. 5.4, where we place a unit

charge with position vector r and let its inverse point with respect to the sphere be P� (seeFig. 5.4) such that

2a′⋅ =OP OP



Let , , ,� � �� OP r OP OQ r� then 2/ .OP a r′ = If Q� be a variable point on the surface of

the sphere, then from the similarity of triangles and ,OQ P OQ P� � � we have

PQ r a

P Q a �

�

� ��

Fig. 5.4 Inverse image point in a sphere.

from which we obtain

�� r

PQ P Qa

This relation is valid for all points on the spherical surface. Therefore, the harmonic frunctionis

1( , )

| | | | | |

a a aH r r

r r r′ = − = − =

′ ′ ′ ′ ′− −P Q OP OQ r�

which can also be written as

2 2

2

( , )a a

Ha a

r rr r r

� ��

� ��

r rr

r r r (5.52)

This form of H satisfies the Laplace equation (5.50). Let ( , , )Q r � �� be a variable point

inside the sphere. When Q lies on the surface of the sphere, say ,Q� we can verify that

PQ r

P Q a�

�(5.53)



Thus the Green’s function to the present problem is

2

2

1 /( , )

| |

a rG

a

r

′ = −′−

′−r r

r rr r (5.54)

or

1 / 1 /( , )

| | | |

a r a rG

R R′ = − = −

′ ′r r

PQ P Q(5.55)

where , .PQ R P Q R� �� On the surface of the sphere, using Eq. (5.53) we can verify that

G vanishes, and hence G defined by Eq. (5.54) is the appropriate Green’s function.Using cosine law in solid geometry, we get

2 2 2 2( ) ( ) 2 cosPQ r r rr R��

2 2 2 2( ) ( ) ( ) 2 ( ) cos ( )P Q OP r OP r R�� (5.56)

or

4 22 2 2

2

2( ) ( ) cos ( )

a aP Q r r R

rrθ′ ′ ′ ′= + − =

From Eq. (5.55), on the sphere,

3

2 2 3 3

1 / 1

( ) ( )

G G R a r R R a R RR R

n r r r r r rR R R R

⎡ ⎤′ ′⎛ ⎞ ′= = − + = − −⎢ ⎥⎜ ⎟′ ′ ′ ′ ′⎝ ⎠′ ′⎢ ⎥⎣ ⎦

� � � � � ��

But, Eq. (5.53) gives

R r

R a�

�

Therefore,

3

3 3

1G R a r RR R

n r r rR a

⎡ ⎤′⎛ ⎞ ′= − −⎢ ⎥⎜ ⎟′ ′⎝ ⎠⎣ ⎦� � ��

Also, Eq. (5.56) yields

2

2 2 2 cos

2 2 2 cos

RR r r

r

R aR r

r r

θ

θ

′= −′

′′ ′= −′

��

��



Hence,2 2

3 2

2 2 2

3 3

1( cos ) cos

1

r a

G r aa r a

n rR a

r r aa

aR aR

θ θ′=

⎡ ⎤⎛ ⎞= − − − −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

⎛ ⎞ −= − − =⎜ ⎟⎝ ⎠

��

or

2 2

2 2 3/2( 2 cos )r a

G r a

n a r a ar θ′=

−=+ −

��

If ( , )u f � �� on the surface of the sphere, then the solution to the interior Dirichlet’s

problem for a sphere is

2 2

2 2 3/2

1 ( , ) ( 1) ( )( , , )

4 ( 2 cos )S

f a ru r dS

a r a ar

θ φθ φπ θ

′ ′− − − ′=+ −∫∫

But, 2 sin .dS a d d� � �� Therefore,

2 2 2

2 2 3/20 0

( ) ( , )sin( , , )

4 ( 2 cos )

a a r f d du r

r a ar

π π θ φ θ θ φθ φπ θ

′ ′ ′ ′ ′−=+ −∫ ∫ (5.57)

which is called the Poisson integral formula.

EXAMPLE 5.4 Consider the case when IR consists of the half-plane defined by 0,x �

,y�� and hence solve 2 0u∇ = in the above region subject to the condition

( ) on 0,u f y x� � using the Green’s function technique.

Solution In this problem 0x � is the boundary. Let ( , )P x y� � be the image point of

( , ).P x y If ( , )Q x y� � is a point on the boundary 0x � (see Fig. 5.5), then ,PQ P Q�� andwe construct the Green’s function G such that

1 1ln ln

| |G H H

PQ� � � �

��r r




Let

1 1ln ln

| |H

P Q� � � �

� �� r�

Then, ln ( / ).G P Q PQ′=

Obviously, 2 0H∇ = in the region 0x � and on the boundary 0, ln 1 0x G= = = is satisfied.Hence the required Green’s function is given by

2 2

2 2

1 ( ) ( )( , ; , ) ln

2 ( ) ( )

x x y yG x y x y

x x y y

� ��

� ��

Here, the outward drawn normal to the boundary is in the direction of the x-axis. Therefore,

2 2 2 2 2 20 0

1 2 2 2

2 ( ) ( ) ( ) ( ) ( )x x

G G x x x

n x x x y y x x y y x y y′= ′=

⎛ ⎞ ⎡ ⎤= − = − + = −⎜ ⎟ ⎢ ⎥′⎝ ⎠ ′ ′ ′ ′ ′+ + − − + − + −⎣ ⎦

� ��

From Eq. (5.39),

2 2

1 2( , ) ( )

2 ( )

xu x y f y dy

x y y�

��

or

2 2

( )( , )

( )�

�

��

�� x f yu x y dy

x y y (5.58)

EXAMPLE 5.5 Determine the Green’s function for the Dirichlet problem for a circle given by

2 0,

( ) on

u r a

u f r aθ∇ = <

= =

Solution Let ( , )P r � and ( , )Q r �� have position vectors and ,�r r and let P� be the

inverse of P with respect to the circle so that 2 andOP OP a P� �� has coordinates 2( / , ),a r θ asshown in Fig. 5.6.

Fig. 5.6 An illustration of Example 5.5



Now, we construct the Green’s function G such that

1ln

| |G H� �

��r r

Let ln ( / )H r P Q a′= ⋅ (as for the sphere) so that it can be verified that

2 0H∇ =

lnr P Q

Ga PQ

��

�(5.59)

On the circle r = a,

ln ln ln 1 0( / )

P Q P QG

PQ r a P Q

′ ′= = = =

′

However,

2 2 2 2 cos ( )PQ r r rr � �� (5.60)

4 22 2

22 cos ( )

a aP Q r r

rrθ θ′ ′ ′ ′= + − − (5.61)

Substituting Eqs. (5.60) and (5.61) into Eq. (5.59), G can be written as

2 2 4 2 2

2 2 2

2 2 2 2

2 2

1 { / 2 cos ( )/ }ln

2 { 2 cos ( )}

1 / 2 cos ( )ln

2 2 cos ( )

r r a r r a rG

a r r rr

a r r a rr

r r rr

θ θθ θ

θ θθ θ

⎡ ⎤′ ′ ′+ − −= ⎢ ⎥′ ′ ′+ − −⎢ ⎥⎣ ⎦

′ ′ ′+ − −=′ ′ ′+ − −

(5.62)

But, on the circle ,r a�

2 2

2 2

( )

[ 2 cos ( ) ]r ar a

G G a r

n r a a ar rθ θ′′ ==

− −⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ′⎝ ⎠⎝ ⎠ ′− − +� ��

Therefore,

2 2 2

2 20

( )( , )

2 { 2 cos ( ) }

a r f du r

a a ar r

π θ θθπ θ θ

′ ′−=′− − +∫

��

Let us consider the Dirichlet boundary value problem described by

2u f∇ = (5.63)



valid in certain region IR, subject to the boundary condition

u g� (5.64)

on IR,� the boundary of IR.

From the definition of Green’s function which has been introduced in Section 5.1. The Green’sfunction must satisfy the relations

2 ( , ) in IRG x yδ ξ η∇ = − − (5.65)

0, on IRG = � (5.66)

Now, consider the eigenvalue problem associated with the operator 2∇ in the domain

IR, i.e.

2 0 in IRφ λφ∇ + = (5.67)

0 on IR� �� (5.68)

Let mn� be the eigenvalues and mn� be the corresponding eigenfunctions. Suppose we give

Fourier series expansion to G and � in terms of the eigenfunctions mn� in the following

form:

( , ; , ) ( , ) ( , )mn mnm n

G x y a x y� � � � �� (5.69)

( , ) ( , ) ( , )mn mnm n

x y b x y� � � � � �� (5.70)

where

2 2IR

( , )1( , ) ( , )

|| || || ||mn

mn mnmn mn

b x y x y dx dyφ ξ ηδ ξ η φ

φ φ= − − =∫∫ (5.71)

2 2

IR

|| ||mn mn dx dyφ φ= ∫∫Now, substituting Eqs. (5.69) and (5.70) into Eqs. (5.65) and (5.66) and noting that Eq. (5.67)has the form

2 0mn mn mnφ λ φ∇ + = (5.72)

we obtain

2 ( , ) ( , ) ( , ) ( , )mn mn mn mnm n m n

a x y b x yξ η φ ξ η φ∇ =∑∑ ∑∑



Using Eqs. (5.71) and (5.72), the above equation reduces to

2

( , ) ( , )( , ) ( , )

|| ||mn mn

mn mn mnm n m n mn

x ya x y

φ ξ η φλ ξ η φφ

− =∑ ∑ ∑ ∑from which we get

2

( , )( , )

|| ||mn

mnmn mn

aφ ξ ηξ η

λ φ= − (5.73)

Hence, Eqs. (5.69) and (5.73) give the required Green’s function for the Dirichlet problem,in the form

2

( , ) ( , )

( , ; , )|| ||

mn mnm n

mn mn

x y

G x y

φ ξ η φξ η

λ φ= −

∑∑(5.74)

EXAMPLE 5.6 Find the Green’s function for the Dirichlet problem on the rectangle

IR:0 ,x a≤ ≤ 0 ,y b� � described by the PDE

2( ) 0 in IRuλ∇ + = (5.75)

and the BC 0 on IR .u = �

Solution The eigenfunctions of the given PDE can be obtained easily by using thevariables separable method. Let us assume the solution of the given PDE in the form

( , ) ( ) ( ).u x y X x Y y= Substituting into the given PDE, we obtain

X Y

X Yλ ν

′′ ′′⎛ ⎞= − + = −⎜ ⎟⎝ ⎠(a separation constant) (5.76)

Since u is zero on the boundary IR, X� satisfies

0, (0) ( ) 0X X X X aν′′ + = = = (5.77)

Its solution, in general, is

( ) cos sinX x A x B xν ν= +

(0) 0X = implies 0.A � ( ) 0X a � gives sin 0,aν = implying / .n aν π= The corresponding

real valued eigenfunctions are sin ( / ), 1, 2, ,nX n x a nπ= = … while the eigenvalues

are 2 2 2/ ,n n a nν π= 1, 2,= … Now, the factor Y satisfies

( ) 0, (0) ( ) 0nY Y Y Y bλ ν′′ + − = = =



Following the above procedure, we can show at once that the eigenfunctions are given by

sin , 1, 2,mm yY m

bπ= = …

and the corresponding eigenvalues are

2 2

2, 1, 2,n

m mbπλ ν− = = …

Thus, we obtain the eigenfunctions to the given problem in the form

( , ) sin sin , 1, 2, ...; 1, 2, ...mnm x n yx y m n

a bπ πφ = = = (5.78)

while the eigenvalues are given by

2 2 2 2 2 22

2 2 2 2mnm n m n

a b a bπ πλ π

= + = +

(5.79)

Computation of || ||mnφ gives

2 2 20 0

|| || sin sin4

a bmn

m x n y abdx dya bp pf = =Ú Ú (5.80)

Hence, the Green’s function for the given Dirichlet problem can be obtained from Eq. (5.74)with the help of Eqs. (5.79) and (5.80) as

2 2 2 2 21 1

sin ( / ) sin ( / ) sin ( / ) sin ( )4( , ; , )m n

m x a n y b m a n babG x ym b n a

=+

p p px phx hp

• •

= =- Â Â (5.81)

5.5 GREEN’S FUNCTION FOR THE WAVE EQUATION—HELMHOLTZTHEOREM

In finding the solution of the wave equation by the variables separable method, we have observedthat the function depending on spatial coordinates satisfies the Helmholtz equation which is alsocalled the spatial form of the wave equation. The solution of the Helmholtz equation under certainboundary conditions can be made to depend on how the appropriate Green’s function is determined,in terms of which the solution to the wave equation can be obtained.

Let u be a solution of the Helmholtz equation

2 2 0u k u∇ + = (5.82)

in the region IR . Also, let all the singularities of u lie outside the closed region IR (see

Fig. 5.7), the boundary of which is denoted by IR .∂ Consider the singularity solution of theHelmholtz equation given by

exp{ | |}/ | |u ik′ ′ ′= − −r r r r (5.83)



Fig. 5.7 A closed region.

Recalling the Green’s theorem (2.19), we have

2 2

IR IR

( ) u uu u u u dV u u dSn n′⎛ ⎞′ ′ ′∇ − ∇ = −⎜ ⎟

⎝ ⎠∫∫∫ ∫∫∂

∂ ∂∂ ∂

(5.84)

where n is the outward normal to IR.∂ Since ( )P r lines outside IR, | | 0,′− ≠r r and it is

possible to calculate 2u ′∇ for all .R′∈r Thus, setting andu uψ ψ′ ′= = into the left-hand sideof Eq. (5.84), we obtain, after using Eqs. (5.82) and (5.83), relation

2 2 2 2

IR IR

2 2

IR

exp ( | |) exp ( | |)( ) ( )| | | |

exp ( | |) exp ( | |)( ) 0| | | |

ik ikdV k dV

ik ikik k dV

ψ ψ ψ ψ ψ ψ

ψ ψ

′ ′⎡ − − ⎤⎧ ⎫′ ′∇ − ∇ = ∇ − −⎨ ⎬⎢ ⎥′ ′− −⎩ ⎭⎣ ⎦

′ ′− −⎡ ⎤= + =⎢ ⎥′ ′− −⎣ ⎦

∫∫∫ ∫∫∫

∫∫∫

r r r rr r r r

r r r rr r r r

Therefore,

IR

exp ( | |) exp ( | |)( ) ( ) 0| | | |ik ik dS

n nψ ψ

′ ′⎡ − − ⎤⎧ ⎫′ ′− =⎨ ⎬⎢ ⎥′ ′− −⎩ ⎭⎣ ⎦∫∫∫

r r r rr rr r r r

∂

∂ ∂∂ ∂

(5.85)

If ( )P r lies inside IR, we surround P by a sphere of radius .ε Now applying Green’s theorem to

the region Σ bounded externally by IR∂ and internally by C as in Fig. 5.1, and noting

that | | 0,′− ≠r r we get

IR

exp ( | |) exp ( | |)( ) ( ) 0| | | |C

ik ik dSn n

ψ ψ′ ′⎧ − − ⎫⎡ ⎤ ⎛ ⎞+ ′ ′− =⎨ ⎬⎢ ⎥ ⎜ ⎟′ ′− −⎣ ⎦ ⎝ ⎠⎩ ⎭

∫∫ ∫∫ r r r rr rr r r r∂

∂ ∂∂ ∂



However,

2

exp ( | |) exp ( ) exp ( ) exp ( ) exp ( ) 1

| |

1 exp ( | |)

| | | |

ik ik ik ik ik ikik

n

ikik

ε ε ε εε ε ε ε εε

′−⎧ ⎫ ⎧ ⎫ ⎛ ⎞= = − = −⎨ ⎬ ⎜ ⎟⎨ ⎬′− ⎝ ⎠⎩ ⎭⎩ ⎭

′−⎛ ⎞= −⎜ ⎟′ ′− −⎝ ⎠

r rr r

r rr r r r

� ��

Hence,

IR

exp ( | |) exp ( | |)( ) ( )

| | | |

1 exp ( | |)( ) ( )

| | | |C

ik ikdS

n n

ikik dS

n

ψ ψ

ψ ψ

′ ′⎡ − − ⎤⎧ ⎫′ ′−⎨ ⎬⎢ ⎥′ ′− −⎩ ⎭⎣ ⎦

′⎧ ⎫ −⎛ ⎞ ′ ′= − − − ⋅⎨ ⎬⎜ ⎟′ ′− −⎝ ⎠⎩ ⎭

∫∫

∫∫

r r r rr r

r r r r

r rr r

r r r r

�

� ��

��

(7.86)

Now using the relations

2

( ) ( ) 0 ( )

sin

0 ( )P

dS d d

n n

ψ ψ ε

ε θ θ φ

ψ ψ ε

′ = +

=

⎛ ⎞= +⎜ ⎟⎝ ⎠

r r

� ��

Equation (5.86) can be rewritten as

21 exp ( | |)[ ( ) 0 ( )] 0 ( ) sin

| | | |PC

ikik d d

n

ψψ ε ε ε θ θ φ′⎧ ⎫ −⎛ ⎞ ⎛ ⎞− + − +⎜ ⎟⎨ ⎬⎜ ⎟′ ′− −⎝ ⎠⎝ ⎠⎩ ⎭∫∫ r r

rr r r r

��

Taking the limit as 0,� � we get

( ) sin 4 ( )C

d dψ θ θ φ πψ− = −∫∫ r r

Hence,

IR

exp ( | |) exp ( | |)( ) ( ) 4 ( )

| | | |

ik ikdS

n nψ ψ πψ

′ ′⎡ − − ⎤⎧ ⎫′ ′− = −⎨ ⎬⎢ ⎥′ ′− −⎩ ⎭⎣ ⎦∫∫ r r r rr r r

r r r r�

� ��

(5.87)

Thus, combining the results (5.85) and (5.87), we have the Helmholtz theorem which

states that if ( )� r is a solution of the spatial form of the wave equation 2 2 0,kψ ψ∇ + =

possessing continuous first and second order partial derivatives in IR bounded by IR,� then



IR

( ) if1 exp ( | |) ( ) exp ( | |)( )

4 | | | | 0 if

Rik ikdS

n n R

ψψ ψπ

′ ∈⎧′ ′ ′⎧ − − ⎫ ⎪⎛ ⎞′− = ⎨⎨ ⎬⎜ ⎟′ ′− −⎝ ⎠ ∉⎩ ⎭ ⎪⎩∫∫

r rr r r r rr

r r r r r�

� ��

��

From this result, it appears as though the values of and / nψ ψ� � on the surface IR� bounding

the region IR have to be prescribed; it also appears that these values can be consideredarbitrarily and independently of each other. But, it can be seen that this is not true if we

introduce the Green’s function ( , ).G �r r Let ( , )G �r r be a Green’s function such that

(i) 2 2( , ) ( , ) 0,G k G′ ′∇ + =r r r r and

(ii) G is finite and continuous with respect to both the variables and .′r r

If we replace exp ( | ) |ik ′ ′− −r r | r r | by G, we get from Eq. (5.88) the relations

IR

1 ( )( ) ( , ) ( ) ( , )

4

GG dS

n n

��

�

��

rr r r r r r

�

� ��

where n is the outward drawn normal to IR.� If 1G G� such that 1G satisfies (i) and (ii) and

1( , ) 0G ′ =r r on IR,� then

IR

1( ) ( ) ( , )

4

GdS

nψ ψ

π′ ′= − ∫∫r r r r

�

��

(5.89)

EXAMPLE 5.7 Determine the Green’s function for the Helmholtz equation for the half-space

0.z �

Solution Here the boundary is the xoy-plane. Let ( )P r be a point and let P� be its

image in the plane 0z � (see Fig. 5.8). Also, let = ( , , );x y zr then ( , , );x y z= −� again, let

( , , ).x y z� � � ��r When �r lies on the boundary, i.e., on the xy plane, ( , , 0)x y�� r and

| | | | .� � �� P r r r

Let

exp ( | |) exp ( | |)( , )

| | | |

ik ikG

� ��

� ��

r r rr r

r r r�

�(5.90)

If �r lies on the boundary, then

on the -planexy

G G

n z⎛ ⎞= −⎜ ⎟′⎝ ⎠

� ��



Fig. 5.8 Illustration of Example 5.7.

But

2 2 2

2 2 2

| | ( ) ( ) ( )

| | ( ) ( ) ( )

x x y y z z

x x y y z z

� � � ��

� � � ��

r r

r�

Therefore,

3 2 3 20

2

2 12

ikR ikR ikR ikR

z

ikR ikR

G ze ikze ze ikze

z R R R R

ze eik

R z RR

′=

⎡ ⎤= − + − +⎢ ⎥′ ⎣ ⎦

⎛ ⎞⎛ ⎞= − = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

��

��

where 2 2 2 2( ) ( )R x x y y z� ��

From the result (5.89), the required Green’s function is

1( ) ( , )

2

ikRef x y dx dy

z Rψ

π� �

�� r

��

(5.91)

where ( , ),f x y� � is the value of � on the boundary 0.z �

EXAMPLE 5.8 Solve the following one-dimensional wave equation using the Green’s functionmethod:

2: 0, 0 , 0PDE

BCs: (0, ) ( , ) 0 for 0

ICs: ( , 0) ( )

( , 0) ( ), 0

tt xx

t

u c u x L t

u t u L t t

u x f x

u x g x x L

− = ≤ ≤ ≥

= = ≥

=

= ≤ ≤



Solution In Section 4.5, the general solution of the one-dimensional wave equation wasobtained as

1

( , ) sin cos sinn nn

n x n ct n ctu x t A B

L L L

π π π�

�

� � � �� (5.92)

where

0

2( )sin

L

nn x

A f x dxL L

π= ∫

0

2( )sin

L

nn x

B g x dxL L

π= ∫Now, let us define a function ( , ; , )G x t � � as follows:

1

2 1( , ; , ) sin sin sin ( )

n

n x n n cG x t t

c n L L L

π πξ πξ τ τπ

�

�

� � �� (5.93)

It can be shown that these series converge for all values of , , , .x t � � It can also be noted that

G as a function of x satisfies the boundary conditions, i.e.,

(0, , , ) 0, ( , , , ) 0, 0G t G L t t� � � ��

Also,

( , ; , ) 0, 0

( , ; , ) ( , ; , ) for all and

( , ; , ) ( , ; , ) for all and

G x x L

G t x G x t x

G x t G x t t

� � �

� � � � �

� � � � �

� � �

�

� �

Thus, the function G defined by Eq. (5.93) is called the Green’s function of the given IBVP.Substituting the series expression for G and formally interchanging the operations of

summation and integration, we can verify that the series solution (5.92) of the given problemcan be rewritten in terms of Green’s function of the form

0 0( , ) ( , ; , 0) ( ) ( , ; , 0) ( )

L L

tu x t G x t f d G x t g dξ ξ ξ ξ ξ ξ= +∫ ∫

��

Consider the diffusion equation (also known as heat conduction equation) with no sourcespresent:

2uk u

t= ∇�

�(5.94)



subject to the boundary condition

( , ) ( , ),u t t S�� r r r (5.95)

and the initial condition

( , 0) ( ),u f V= ∈r r r (5.96)

It is necessary for us to find ( , )u tr in the volume V bounded by the surface S by using

the Green’s function technique.

We shall define the Green’s function ( , , ), ,G t t t t� � �� r r where t� is a parameter, such

that the following conditions are satisfied:

(i) 2Gk G

t= ∇�

�(5.97)

(ii) BC: ( , , ) 0,G t t S′ ′ ′− = ∈r r r (5.98)

(iii) The initial condition Lt 0t t

G��

� for all points in the volume V except at the point ,r

where G assumes the singularity solution as given in the introduction in the form

2

3/2

1 | |exp

4 ( )8[ ( )] k t tk t tπ

⎡ ⎤′−−⎢ ⎥′−′− ⎣ ⎦

r r (5.99)

It is easy to note that G depends only on t through the term ( )t t�� . Hence, equivalently, Eq. (5.97)

can also be written as

2Gk G

t= − ∇

′��

(5.100)

Here, G can be interpreted as the temperature at the point �r at time t, corresponding to a

source of unit strength generated at .t t�� Initially, the solid with volume V and surface S is

at zero temperature. Since t� must lie within the time interval for t for which Eqs. (5.94) and(5.95) are valid, these equations may be rewritten as

2 ,u

k u t tt

′= − ∇ <′

��

(5.101)

( , ) ( , ),u t t S�� r r r (5.102)

Also,

( )u G

uG G ut t t

= +′ ′ ′

� � ��

Now, using Eqs. (5.100) and (5.101), we have

2 2( ) � � � ��

uG Gk u ku Gt

��



If � is an arbitrary positive constant, then

2 2

0 0( ) ( )

t t

V V

uG d dt k G u u G d dtt

ε ετ τ

− −⎧ ⎫⎪ ⎪ ⎧ ⎫′ ′ ′ ′= ∇ − ∇⎨ ⎬ ⎨ ⎬′⎪ ⎪⎩ ⎭ ⎩ ⎭∫ ∫∫∫ ∫ ∫∫∫�

�(5.103)

Interchanging the order of integration on the left-hand side of the above equation, we have

0 00

( ) ( ) ( ) ( )t t

t t t

V V V V

uG dt d uG d uG d uG dt

ε εετ τ τ τ

− −′ ′= − =

⎧ ⎫′ ′ ′ ′ ′= = −⎨ ⎬′⎩ ⎭∫∫∫ ∫ ∫∫∫ ∫∫∫ ∫∫∫��

But

[ ( , )] ( , )t tu t u tε ε′= −′ ′ ′= −r r

and we want values of u for times greater than some initial time. Hence, ( , )u t �� r can be

taken as ( , )u t�r . After using IC (5.96), the left-hand side of Eq. (5.103) becomes

( , ) ( , , ) ( , , ) ( )t tV V

u t G t t d G t f dε τ τ′= −′ ′ ′ ′ ′ ′− −∫∫∫ ∫∫∫r r r r r r

From the expression (5.99) for ( , , ),G t t� ��r r we can show that

2

3/2

1 | |( , , ) exp 1

48( )

dt t

V V

G t tkk

τε επ ε

′′= −

⎡ ⎤′−′ ′− = − =⎢ ⎥⎣ ⎦∫∫∫ ∫∫∫ r r

r r

as 0.� � After applying Green’s theorem, the right-hand side of Eq. (5.103) beomes

2 2

0 0( )

t t

V S

u Gk G u u G d dt k G u dS dt

n n

� �

�

� � � ��

� ��

But 0 on .G S� Now taking the limit as 0,� � the above equation becomes

0( , )

t

S

Gk t dS dt

nθ⎧ ⎫⎪ ⎪′ ′ ′−⎨ ⎬

⎪ ⎪⎩ ⎭∫ ∫∫ r

��

Finally, Eq. (5.103) reduces to the form

0( , ) ( ) ( , , ) ( , )

t

V S

Gu t f G t d k dt t dS

nτ θ′ ′ ′ ′ ′ ′= −∫∫∫ ∫ ∫∫r r r r r

��

(5.104)

which is the required solution to the boundary value problem described by Eqs. (5.94)–(5.96).



EXAMPLE 5.9 Determine Green’s function for the problem of heat flow in an infinite roddescribed by

PDE: , , 0

IC: ( , 0) ( ),

t xxu u x t

u x f x x

α� ��

� ��

Solution The variables separable method of solution as discussed in Section 3.3 to thegiven problem is

21( , ) ( ) exp { ( ) /(4 )} ( , ) ( )

4u x t f y x y t dy G x y t f y dy

tα α

πα� �

��

where

21( , ) [exp{ ( ) /(4 )}]

4G x y t x y t

tα α

π α− = − − (5.106)

is called the Green’s function for heat transfer in infinite rod.

EXAMPLE 5.10 Find a Green’s function for the heat flow problem in a finite rod describedby

PDE: , 0 , 0

BCs: (0, ) ( , ) 0, 0

IC: ( , 0) ( ), 0

t xxu u x L t

u t u L t t

u x f x x L

α= ≤ ≤ >

= = >

= ≤ ≤

Solution In Example 3.5, we have obtained the variables separable solution to the givenPDE; its general form is

2( , ) sintu x t Ce x��

��

�

Applying the BCs: (0, ) ( , ) 0,u t u L t� � we get

2( , ) sin 0tu L t Ce L��

��

� �

which means that sin 0.L� � Therefore, / , 1, 2, ...n L nλ π= =Using the superposition principle, we obtain

2

1

( , ) exp ( ) sin , /n n n nn

u x t C t x n Lαλ λ λ π�

�� (5.107)

Now, using the IC: ( , 0) ( )u x f x� , we get

1

( ) sinn nn

f x C xλ�

��



which is a half-range Fourier sine series. Therefore,

0

2( ) sin

L

n nC f y y dyL

λ= ∫ (5.108)

Inserting Eq. (5.108) into Eq. (5.107), we obtain the following integral representation of thesolution:

2

01

2( , ) exp ( ) sin sin ( )

L

n n nn

u x t t y x f y dyL

αλ λ λ�

�

� ��

� � �� (5.109)

If we define

2

1

2( , , ) exp ( ) sin sinn n n

n

G x y t t y xL

αλ λ λ�

�� (5.110)

for 0 , and 0,x y L t� � � then solution (5.109) can be expressed as

0( , ) ( , , ) ( ) for 0 , 0

Lu x t G x y t f y dy x L t= ≤ ≤ >∫ (5.111)

The function ( , , )G x y t defined in Eq. (5.110) is called the Green’s function for the given

heat equation.It can be observed that this function has the following properties:

(i)

(ii)

(iii)

, ,

,

,

( , , ) ( , ) ( , )

(0, , ) ( , ) 0

( , , ) ( , )

t xx yyt t

t

t

G x y t G x y G x y

G y t G L y

G x y t G y x

α α= =

= =

=

for 0 , , 0

for 0 , 0

for 0 , , 0

x y L t

y L t

x y L t

≤ ≤ >

≤ ≤ >

≤ ≤ >

(5.112)

(5.113)

(5.114)

Equation (5.112) follows from Eq. (5.110) by term-by-term differentiation. Thus the Green’sfunction satisfies the heat equation. In fact, it is symmetric and satisfies the boundary conditions.

�!��

1. Show that the three-dimensional Dirac -functionδ can be written as

( , , ; , , ) ( ) ( ) ( )x y x yδ ξ η ζ δ ξ δ η δ ζ= − − −z z

2. Let 0 0( , )P x y be a point in rectangular coordinates corresponding to the point 0 0( , )P r �

in polar coordinates. Then show that

0 0 0 0( ) ( ) ( ) ( )x x y y r r rδ δ δ δ θ θ− − = − −



3. Let 1 1 1( , , )P x y z be a point in rectangular coordinates corresponding to the point

1 1 1( , , )OP r � � in spherical polar coordinates. Then, show that

1 1 11 1 1 2

( ) ( ) ( )( ) ( ) ( )

sin

r rx x y y z z

r

δ δ θ θ δ φ φδ δ δθ

− − −− − − =

4. Use the method of images and show that the harmonic Green’s function for the half-space 0z ≥ is

1 1( , )

4 4G

r r� �

� � ��

r r

where

2 2 2 21 1 1( ) ( ) ( )r x x y y z z= − + − + −

2 2 2 21 1 1( ) ( ) ( )r x x y y z z′ = − + − + +

5. Solve

2 0u∇ =

in the upper half-plane defined by 0, ,y x� �� using Green’s function method,subject to the condition

( ) on 0u f x y� �

6. Determine the Green’s function for the Robin’s problem on the quarter infinite planedescribed by

2 ( , ), 0, 0u x y x yφ∇ = > >

subject to the conditions

( ) on 0

( ) on 0

u f y x

ug x y

n

��

� �

� �

7. Show that the Green’s function for the heat flow problem in semi-infinite rod describedby

PDE: , 0, 0

BC: (0, ) 0, 0

IC: ( , 0) ( ), 0

t xxu u x t

u t t

u x f x x

α� � �

� �

� �

is given in the form

2 21( , , ) [exp { ( ) /4 } exp { ( ) /4 }]

4G x y t x y t x y t

tπ� � � � � �


��

Laplace transform is essentially a mathematical tool which can be used to solve severalproblems in science and engineering. This transform was first introduced by Laplace, aFrench mathematician, in the year 1790 in his work on probability theorem. This techniquebecame popular when Heaviside applied to the solution of an ordinary differential equationreferred hereafter as ODE, representing a problem in electrical engineering. To the basicquestion as to why one should learn Laplace transform technique when other techinques areavailable, the answer is very simple. Transforms are used to accomplish the solution of certainproblems with less effort and in a simple routine way. To illustrate, consider the problem offinding the value of x from the equation

1.85 3x �(6.1)

It is an extremely tedious task to solve this problem algebraically. However, taking logarithmson both sides, we have the transformed equation as

1.85 ln ln 3x � (6.2)

In this transformed equation, the algebraic operation and exponentiation have been changedto multiplication which immediately gives

ln 3ln

1.85x �

To get the required result, it is enough if we take the antilogarithm on both sides of the aboveequation, which yields

1 ln 3ln

1.85x − ⎛ ⎞= ⎜ ⎟⎝ ⎠

With the help of any ordinary calculator, we can now compute x. Following this simpleexample, the Laplace transform method reduces the solution of an ODE to the solution of an

316

��

Laplace Transform Methods


LAPLACE TRANSFORM METHODS 317

algebraic equation. In fact, this method has a particular advantage in finding the solution ofan ODE with appropriate ICs, without first finding the general solution and then using ICsfor evaluating the arbitrary constants. Also, when the Laplace transform technique is appliedto a PDE, it reduces the number of independent variables by one.

Definition 6.1 Suppose ( )f t is a piecewise continuous function and if it has an additional

property that there exists a real number 0� and a finite positive number M such that

0Lt | ( ) | fort

tf t e M�

� ��

�

� ��

and the limit does not exist when 0 ,γ γ< then such a function is said to be of exponential

order 0 ,� also written as

0| ( ) | 0( )tf t e��

Variables such as velocity and current are always finite; which means that ( )f t is bounded.

Thus for any bounded function ( ), | ( ) | 0tf t f t e �� for all 0.� � The order of such a function

is zero. However, variables such as electrical charge and mechanical displacement may increasewithout limit but of course proportional to t. Such functions are also of exponential order.

For illustration, let us consider the following examples:

(i) Lt 0t

tte ��

�

�

�

The fact that nt is of exponential order zero can be seen as follows:

1

Lt Lt Ltn n

n tt tt t t

t ntt e

e e

�

� ��

�

�

� � �

� � � ��

� � � �� (using L’Hospital’s rule)

Applying the L’Hospital’s rule repeatedly, we get

!Lt Lt 0n t

n tt t

nt e

e�

��

�

� �

� ��

� ��

(ii) In an unstable system a function may increase as eat and we can see that

Lt 0 ifat t

te e a�

��

�

� �

�

Thus the function eat is of exponential order a.

(iii) exp ( ) ( 1)nt n � is not of exponential order, since

1Lt exp ( ) Lt exp [ ( )]n t n

t tt e t t�

��

� �

� � �

� ��

for any finite value of .�



Definition 6.2 Let ( )f t be a continuous and single-valued function of the real variable t

defined for all , 0 ,t t� � � and is of exponential order. Then the Laplace transform of ( )f t

is defined as a function ( )F s denoted by the integral

0[ ( ); ] ( ) ( )stL f t s F s e f t dt�

� � ��

(6.3)

over that range of values of s for which the integral exists. Here, s is a parameter, real or

complex. Obviously, [ ( ); ]L f t s is a function of s. Thus,

1

[ ( ); ] ( )

( ) [ ( ); ]

L f t s F s

f t L F s t−

=

=

where L is the operator which transforms ( ) into ( )f t F s , called Laplace transform operator,

and 1L� is the inverse Laplace transform operator.The Laplace transform belongs to the family of “integral transforms’’. An integral transform

( )F s of the function f (t) is defined by an integral of the form

( , ) ( ) ( )b

ak s t f t dt F s=∫ (6.4)

where ( , ),k s t a function of two variables s and t, is called the kernel of the integral transform.

The kernels and limits of integration for various integral transforms are given in Table 6.1(which is not exhaustive).

Table 6.1 Kernels and Limits for Various Integral Transforms

Name of the transform ( , )k s t a b

Laplace transform ste� 0 �

Fourier transform / 2iste π ��

Fourier sine transform2

sin st�

0 �

Fourier cosine transform2

cos st�

0 �

Hankel transform ( )ntJ st 0 �

Mellin transform 1st � 0 �

The integral transforms defined above are applicable, either for semi-infinite or infinitedomains. Similarly, finite integral transforms can be defined on finite domains.

Now, we are in a position to verify the following important result.



Theorem 6.1 If ( )f t is piecewise continuous in the range 0t � and is of exponential order

,� then the Laplace transform ( ) of ( )F s f t exists for all .s ��

Proof From the definition of Laplace transform,

1 20 0

[ ( ); ] ( ) ( ) ( )Tst st st

TL f t s e f t dt e f t dt e f t dt I I� � �

� � � � ��

Since ( )f t is piecewise continuous on every finite interval 10 ,t T I� � exists, whereas

2| | | ( ) |st

TI e f t dt�

� ��

But ( )f t is a function of exponential order; therefore,

| ( ) | for realtf t Me� ��

Hence,( )| ( ) |st s te f t Me ��

�

Thus,

( )( )

2| | ,s T

s t

T

MeI e M dt s

s

��

��

� �

� �

� � �

��

In other words, 2I can be made as small as we like provided T is large enough and, therefore,

2I exists. Hence, [ ( ); ]L f t s exists for .s ��

��

Following the definition of Laplace transform by the integral (6.3), we shall compute theLaplace transform of some elementary functions.

EXAMPLE 6.1 Find the Laplace transform of

(i) 1, (ii) 0, (iii) t, (iv) eat, (v) e–at.

Solution Using the definition of Laplace transform, we have

(i)0

0

1[1; ] (1) if 0

stst e

L s e dt ss s

�

�

� ��

��

(ii)0

[0; ] (0) 0stL s e dt�

� ��

(iii)20 0 0

0

1[ ; ]

st st stst e e e

L t s e t dt td t dts s s s

� � �

�

� � � ��

��



(iv)( )

( )

0 00

1[ ; ] ,

( )

s a tat st at s a t e

L e s e e dt e dt s as a s a

� �

� � �

� ��

� � ��

��

(v) ( )

0 0

1[ ; ]at st at s a tL e s e e dt e dt

s a� � � � �

� � �

��


(i) cos at, (ii) sin at.

Solution Following the definition of Laplace transform, we have

(i)0 0

[cos ; ] cos Re Re [ ; ]st st iat iatL at s e at dt e e dt L e s� �

� � ��

2 2 2 2

1Re Re

s ia s

s ia s a s a

+= =− + +

(ii)2 2 2 2

[sin ; ] Im [ ; ] Imiat s ia aL at s L e s

s a s a

+= = =+ +


(i) cosh at, (ii) sinh at.

Solution Using the results established in Example 6.1, we have

(i)

2 2

1[cosh ; ] ; { [ ; ] [ ; ]}

2 2

1 1 1

2

at atat ate e

L at s L s L e s L e s

s

s a s a s a

−−⎡ ⎤+= = +⎢ ⎥

⎢ ⎥⎣ ⎦

⎛ ⎞= + =⎜ ⎟− +⎝ ⎠ −

(ii)

2 2

1[sinh ; ] ; { [ ; ] [ ; ]}

2 2

1 1 1

2

at atat ate e

L at s L s L e s L e s

a

s a s a s a

−−⎡ ⎤−= = −⎢ ⎥

⎢ ⎥⎣ ⎦

⎛ ⎞= − =⎜ ⎟− +⎝ ⎠ −

EXAMPLE 6.4 Find the Laplace transform of ,nt where n is a positive integer.


1 1

0 0 0 00

[ ; ]st st

n st n n n n st n ste e n nL t s e t dt t d t t e dt t e dt

s s s s

� �

� � � � �

� � � ��

��



Hence,

1[ ; ] [ ; ]n nnL t s L t s

s�

�

Similarly, we can prove the following:

1 21[ ; ] [ ; ]n nn

L t s L t ss

� �

�

�

2 32[ ; ] [ ; ]n nn

L t s L t ss

− −−=

�

2

2

2[ ; ] [ ; ]

1[ ; ]

L t s L t ss

L t ss

=

=

Therefore,

2 1

1 2 2 1 ![ ; ] ...n

n

n n n nL t s

s s s s s s �

� �

� � � � �

which can be expressed in Gamma function as

1

1[ ; ]n

n

nL t s

s �

��

��

We present a few important properties of the Laplace transform in the following theoremswhich will enable us to find the Laplace transform of a combination of functions whosetransforms are known.

Theorem 6.2 (Linearity property). If 1 2andc c are any two constants and if 1 2( ) and ( )F s F s

are the Laplace transforms, respectively of 1 2( ) and ( ),f t f t then

1 1 2 2 1 1 2 2 1 1 2 2[{ ( ) ( )}; ] [ ( ); ] [ ( ); ] ( ) ( )L c f t c f t s c L f t s c L f t s c F s c F s� � � � �

Proof Following the definition of Laplace transform, we have

1 1 2 2 1 1 2 20

1 1 2 20 0

1 1 2 2

1 1 2 2

[{ ( ) ( )}; ] { ( ) ( )}

( ) ( )

[ ( ); ] [ ( ); ]

( ) ( )

st

st st

L c f t c f t s e c f t c f t dt

c e f t dt c e f t dt

c L f t s c L f t s

c F s c F s

�

� �

� � �

� �

� �

� �

�

� �

�

� �



Theorem 6.3 (Shifting property). If a function is multiplied by eat, the transform of the

resultant is obtained by replacing s by ( )s a� in the transform of the original function. That

is, if

[ ( ); ] ( )L f t s F s�

then

[ ( ); ] ( )atL e f t s F s a� �

Proof From the definition of Laplace transform,

( )

0 0[ ( ); ] ( ) ( ) ( )at st at s a tL e f t s e e f t dt e f t dt F s a� � �

� � � ��

Similarly,

[ ( ); ] ( )atL e f t s F s a�

� � (6.5)

Theorem 6.4 (Multiplication by power of t). If

[ ( ); ] ( )L f t s F s�

then

( )[ ( ); ] ( 1) ( ) ( 1) ( )n

n n n nn

dL t f t s F s F s

ds� � � �

where 1, 2, 3,n = …

Proof From the definition of Laplace transform

0( ) [ ( ); ] ( )stF s L f t s e f t dt�

� � ��

Hence,

0[ ( )] ( )std dF s e f t dt

ds ds−⎡ ⎤= ⎢ ⎥⎣ ⎦∫

�

Interchanging the operations of differentiation and integration for which we assume that thenecessary conditions are satisfied, and since there are two variables s and t, we use the notationof partial differentiation and obtain

0 0[ ( )] { ( )} ( ) [ ( ); ]st stdF s e f t dt e tf t dt L tf t s

ds s− −= = − = −∫ ∫

� ��

Therefore,

[ ( ); ] ( )d

L tf t s F sds

� �



By repeated application of the above result, it can be shown that

( )[ ( ); ] ( 1) ( ) ( 1) ( )n

n n n nn

dL t f t s F s F sds

= − = − (6.6)

Theorem 6.5 (Differentiation property). If

[ ( ); ] ( )L f t s F s=

then

( ) 1 2 2 1[ ( ); ] ( ) (0) (0) (0) (0)Ln n n n n nL f t s s F s s f s f sf f− − − −= − − − − −′

Proof From the definition of Laplace transform, we have

00 0[ ( ); ] ( ) [ ( )] ( )

(0) [ ( ); ] ( ) (0)

st st stL f t s e f t dt e f t s e f t dt

f sL f t s sF s f

∞ ∞− − ∞ −= = +′ ′

= − + = −

∫ ∫

Similarly, it can be shown that

2

3 2

[ ( ); ] [ ( ); ] (0) { ( ) (0)} (0) ( ) (0) (0)

[ ( ); ] ( ) (0) (0) (0)

L f t s sL f t s f s sF s f f s F s sf f

L f t s s F s s f sf f

= − = − − = − −′′ ′ ′ ′ ′

= − − −′′′ ′ ′′

Thus, in general,

( ) 1 2 3 ( 1)[ ( ); ] = ( ) (0) (0) (0) (0)- - - -- - - - -¢ ¢¢ Ln n n n n nL f t s s F s s f s f s f f (6.7)

This property is very useful for solving differential equations.


(i) cos ,ate bt (ii) sin ,ate bt (iii) cosh ,ate bt (iv) ,at ne t and (v) cos cosh .at bt

Solution Using the shifting property

(i) 2 2 2 2( )

[ cos ; ]( )

at

s s a

s s aL e bt ss b s a b→ −

−= =+ − +

(ii) 2 2 2 2( )

[ sin ; ]( )

at

s s a

b bL e bt ss b s a b→ −

= =+ − +

(iii) 2 2 2 2( )

[ cosh ; ]( )

at

s s a

s s aL e bt ss b s a b→ −

−= =− − −



(iv)1 1

( )

! ![ ; ]

( )at n

n ns s a

n nL e t s

s s a� ��

� ��

(v)

2 2 2 2

[cos cosh ; ] cos ;2

1{ [ cos ; ] [ cos ; ]}

2

1

2 ( ) ( )

bt bt

bt bt

e eL at bt s L at s

L e at s L e at s

s b s b

s b a s b a

�

�

� ��

� �

� ��

EXAMPLE 6.6 Fing the Laplace transform of the following:

(i) 2 ,att e (ii) sin ,t at (iii) 2 cos ,t at (iv) n att e� .

Solution Using the result established in Theorem 6.4, we have

(i)2 2

2 22 2 2 3

1 1 2[ ; ] ( 1) [ ; ]

( ) ( )at atd d d

L t e s L e ss a dsds ds s a s a

��

Alternatively,

23 3

( )

2! 2[ ; ]

( )at

s s a

L e t ss s a� �

� ��

(using the shifting property)

(ii) 12 2 2 2 2

2[ sin ; ] ( 1) [sin ; ]

( )

d d a asL t at s L at s

ds ds s a s a

� ��

(iii)2 2

2 22 2 2 2

2 2 3 2

2 2 2 2 2 3

[ cos ; ] ( 1) [cos ; ]

2 6

( ) ( )

d d sL t at s L at s

ds ds s a

d a s s sa

ds s a s a

� ��

� ��

(iv)1 1

( )

! ![ ; ]

( )at n

n ns s a

n nL e t s

s s a�

� ��

� ��

(using the shifting property)


(i) 4 sin 3 ,tte t� (ii) sin 2 sin 3 ,t t (iii) 3sin 2 .t



Solution We may note that

(i) 4 4[ sin 3 ; ] [ ( sin 3 ); ]t tL te t s L e t t s� ��

Using the result of Theorem 6.4, we get

2 2 2

3 6[ sin 3 ; ]

9 ( 9)

d sL t t s

ds s s

� ��

Now, using the shifting property, we obtain

42 2 2 2 2 2

( 4)

6 6( 4) 6( 4)[ ( sin 3 ); ]

( 9) {( 4) 9} ( 8 25)t

s s

s s sL e t t s

s s s s�

� �

� ��

� � � � �

(ii) Since 1

sin 2 sin 3 (cos cos 5 ),2

t t t t� �

2 2 2 2

1[sin 2 sin 3 ; ] { [cos ; ] [cos 5 ; ]}

2

1 12

2 1 25 ( 1) ( 25)

L t t s L t s L t s

s s s

s s s s

� �

� ��

(iii) Since 3sin 6 sin 3 (2 ) 3 sin 2 4 sin 2 ,t t t t� � � we have 3 3 1

sin 2 sin 2 sin 6 .4 4

t t t� �

Thus,

32 2

2 2

3 1 3 2 1 6[sin 2 ; ] [sin 2 ; ] [sin 6 ; ]

4 4 4 44 36

48

( 4)( 36)

L t s L t s L t ss s

s s

� � � ��

�

EXAMPLE 6.8 Find the Laplace transform of ( )f t defined as

sin , 0( )

0,

t tf t

t

ππ

� ��


( )

0 0 0

( ) ( )

20

2 2

[ ( ); ] sin (0) Im Im

1Im Im Im [1 (cos sin )]

1

1Im [1 ]

1 1

st st st it i s t

i s t i ss

ss

L f t s e t dt e dt e e dt e dt

e e i se i

i s i s s

i s ee

s s

π π π

π

π ππ

ππ

π π

��

� ��

��

� � � �

� � � ��

� ��

� �

� � � �



Theorem 6.6 (Initial value theorem). If ( )f t and ( )f t are Laplace transformable and

( )F s is the Laplace transform of ( )f t , then the behaviour of ( )f t in the neighbourhood of

0t � corresponds to the bebaviour of ( )sF s in the neighbourhood of .s � � Mathematically,

0Lt ( ) Lt ( )

t sf t sF s

� ��

Proof From the property of derivative, we have

[ ( ); ] ( ) (0)L f t s sF s f� � (6.8)

Taking the limit as s �� on both sides, we get

0Lt ( ) Lt ( ) Lt (0)st

s s se f t dt sF s f

� ��

� �� (6.9)

since s is independent of t, we can take the limit before integrating the left-hand side of Eq.(6.9), thus getting

0 0Lt ( ) Lt ( ) 0st st

s se f t dt e f t dt

� ��

� �� and Eq. (6.9) becomes

0Lt ( ) (0) Lt ( )

s tsF s f f t

��

Hence the result. For example, let ( )f t be a polynomial of degree n of the form

20 1 2( ) n

nf t a a t a t a t��

Its Laplace transform is

0 1 22 3 1

!2( ) k

n

a n aa aF s

s s s s� ��

Now, taking the limit on both sides as ,s �� we obtain

0Lt ( ) (0)s

sF s a f��

� �

EXAMPLE 6.9 Verify the initial value theorem for the function

( ) 1 (sin cos )tf t e t t��

Solution Given ( ) 1 (sin cos )tf t e t t�� , we have



2 2( 1)

2

( ) [ ( ); ] [1; ] [( sin cos ); ]

1 1

1 1

1 2

2 2

t t

s s

F s L f t s L s L e t e t s

s

s s s

s

s s s

� �

� �

� � � �

� ��

�� Hence,

2

2

2( ) 1

2 2

s ssF s

s s

��

� �

Therefore,

2

2/ 1Lt ( ) Lt 1 1 1 2

1 2/ 2/s s

ssF s

s s��

��

� �

But (0) 1 1 2.f � � � Thus,

Lt ( ) (0)s

sF s f��

�

Hence the result.

Theorem 6.7 (Final value theorem). If ( ) and ( )f t f t are Laplace transformable and ( )F s

is the Laplace transform of ( ),f t then the behaviour of ( )f t in the neighbourhood of t � �corresponds to the behaviour of ( )sF s in the neighbourhood of 0.s � Mathematically,

0Lt ( ) Lt ( )

t sf t sF s

��

Proof From the property of derivative, we have

[ ( ); ] [ ( ); ] (0)L f t s sL f t s f� �

i.e.

0( ) ( ) (0)ste f t dt sF s f

� � � ��Taking the limit as 0s � on both sides of the above equation, we have

00 0 0Lt ( ) Lt ( ) Lt (0)st

s s se f t dt sF s f

� ��

� ��But,

00 00

Lt ( ) ( ) [ ( )] Lt ( ) (0)st

s te f t dt f t dt f t f t f

� ��

� � � ��



Using this result in the above equation, we get

0Lt ( ) Lt ( )

s tsF s f t

� ��

Theorem 6.8 (Division by t). If

[ ( ); ] ( )L f t s F s�then

( ); ( )

s

f tL s F s ds

t

�� (6.10)

Proof From the definition of Laplace transform

0[ ( ); ] ( ) ( )stL f t s F s e f t dt

� �� Integrating the above equation with respect to s between the limits s to ,� we get

0 0

0 0

( ) ( ) ( ) (by changing the order of integration)

( ) ( )( ) ;

st st

s s s

stst

s

F s ds e f t dt ds f t e ds dt

e f t f tf t dt e dt L s

t t t

� � � � ��

��

� � � ��

� � � ��

� � � � �

� �Hence the result.Note: In applying this rule, one should be careful. Since f (t)/t may have an infinite discontinuity

at 0,t � it may not be integrable. If f (t)/t is not integrable, then its Laplace transform does

not exist. For example, at 0,t � the function sin t/t does not have an infinite discontinuity,while the function cos t/t has an infinite discontinuity.


(i) 1 cos,

t

t

� (ii) cos 2 cos 3t t

t

� .

Solution Using the result of Theorem 6.8, we have

(i) 1 cos; ( )

s

tL s F s ds

t

�� where,

2

1( ) [(1 cos ); ] [1; ] [cos ; ]

1

sF s L t s L s L t s

s s� � � � � �

�



Therefore,

2

22 1/2

2 1/2 2 1/2

2 1/2

1 cos 1;

1

1ln ln ( 1) ln

2 ( 1)

1ln 0 ln

(1 1/ ) ( 1)

( 1)ln

s

s s

s

t sL s ds

t s s

ss s

s

s

s s

s

s

�

� �

�

��

� � � � � � � ��

� � � � ��

��

�

(ii)cos 2 cos 3

; ( )s

t tL s F s ds

t

�� where

2 2( ) [(cos 2 cos 3 ); ]

4 9

s sF s L t t s

s s� � � �

� �Therefore,

2 2

2 2

2 2

2

2

cos 2 cos 3;

4 9

1 4 1 1 4/ln ln

2 29 1 9/

1 9ln

2 4

s

s s

t t s sL s ds

t s s

s s

s s

s

s

�

� �

��

� ��

� ��

�

��

A function f (t) is called periodic with period T, if ( ) ( )f t T f t� � for all values of t and T > 0.

For example, the trigonometric functions sin t and cos t are periodic functions of period

2 .� Periodic functions occur very often in a variety of engineering problems.

Theorem 6.9 If ( )f t is a periodic function with period T, then

0[ ( ); ] ( ) /(1 )

T st stL f t s e f t dt e� �� (6.11)




0 0[ ( ); ] ( ) ( ) ( )

Tst st st

TL f t s e f t dt e f t dt e f t dt

� �� If we substitute t u T� � in the second integral on the right-hand side and

write ,dt du� we obtain

( )

0 0

0 0

0

[ ( ); ] ( ) ( )

( ) ( )

( ) [ ( ); ]

T st s u T

T st sT su

T st sT

L f t s e f t dt e f u T du

e f t dt e e f u du

e f t dt e L f t s

��

��

� �

� � �

� �

� �

� �

� �

�Rearranging, we get

0(1 ) [ ( ); ] ( )

TsT ste L f t s e f t dt� �� Thus,

0[ ( ); ] ( ) /(1 )

T st sTL f t s e f t dt e� �� Hence the result.

EXAMPLE 6.11 Obtain the Laplace transform of the periodic saw-tooth wave functiongiven by

( ) of period , 0t

f t T t TT

� � �

Solution The graph of the periodic saw-tooth function is described in Fig. 6.1. Since

( )f t

is periodic with period T, we have

0 0 0

00

2

1 1 1[ ( ); ]

1 (1 ) (1 )

1 1

(1 )

1 1( 1)

(1 )

stT T Tst stsT sT sT

Tst T stsT

sTsT

sT

t eL f t s e dt e tdt td

T se T e T e

tee dt

s sT e

Tee

sT e s

��

� � �

��

�

��

�

� ��

� ��

� ��

� � �

�



t

f t( )

O

1

T 2T 3T

Fig. 6.1 Saw-tooth function with period T.

Therefore,

2

1[ ( ); ]

(1 )

sT

sT

eL f t s

s T s e

�

��

EXAMPLE 6.12 Find the Laplace transform of the following full wave rectifier function:

sin , 0 /( )

0, / 2 /

E t tf t

t

ω λ ωλ ω λ ω� ��

� � � �

Given that

2( )f t f t

�

�

��

Solution Since the given function f (t) is periodic with period 2 / ,λ ω we have

2 /

2 / 0

/ 2 /

2 / 0 /

/

2 / 2 20

1[ ( ); ] ( )

1

1sin (0)

1

( sin cos )1

sts

st sts

st

s

L f t s e f t dte

e E t dt e dte

E es t t

e s

λ ω

λ ω

λ ω λ ωλ ω λ ω

λ ω

λ ω

ω

ω ω ωω

��

� ��

�

�

��

� � � ��

� � � �

� � �

�

� �

Therefore,

/

2 / 2 2 2 2[ ( ); ] ( sin cos )

1

s

s

E eL f t s s

e s s

λ ω

λ ωωλ ω λ

ω ω

�

�

� ��




1, 0 2( )

1, 2 4

( 4) ( )

tf t

t

f t f t

≤ <⎧= ⎨− ≤ <⎩

+ =

Solution In this problem, ( )f t is a periodic function of period 4; we, therefore, have

4

4 0

2 4

4 0 2

2 4

4

1[ ( ); ] ( )1

1 (1) ( 1)1

1 2 11

sts

st sts

s s

s

L f t s e f t dte

e dt e dte

e es s se

−−

− −−

− −

−

=−

⎛ ⎞= + −⎜ ⎟⎝ ⎠−

⎛ ⎞−= + +⎜ ⎟⎝ ⎠−

∫

∫ ∫

6.5 TRANSFORM OF ERROR FUNCTION

The error function denoted by erf (t) is defined as

2

0

2erf ( )t ut e du

π−= ∫ (6.12)

This function occurs in many branches of science and engineering; for example, in probabilitytheory, the theory of heat conduction, and so on. In terms of the power series, we have

20 0

2 ( 1)erf ( )!

nt n

nt u du

nπ

∞

=

−= ∑∫ (6.13)

Alternatively, it can be written as

2 1

0

2 ( 1)erf ( )! (2 1)

n n

n

ttn nπ

∞ +

=

−=+∑ (6.14)

We can easily verify that these series converge everywhere and, therefore, erf (t) is an entirefunction. From the definition (6.12), it can be verified at once that

erf (0) 0= (6.15)

2

0

2 1/2erf ( ) 1ue duπ π

∞ −∞ = = =∫ (6.16)



The graph of error function is shown in Fig. 6.2.

O

1

erf( )t

t

Fig. 6.2 Error function.

In solving heat conduction equation, it has been found useful to introduce the complementaryerror function defined as

2 2 2

0 0

2 2erfc ( )

tu u u

tt e du e du e du

� �

� �� (6.17)

Therefore,

erfc ( ) 1 erf ( )t t� � (6.18)

Now we shall find the Laplace transform of erf (t): From the definition of Laplacetransform,

2

0 0

2[erf ( ); ]

tst uL t s e e du dt�

� � �� Changing the order of integration, we obtain

2

2

2 2

0 0

( )

0

/4 ( /2)

0

2[erf ( ); ]

2

2

u st

u su

s u s

L t s e e dt du

e dus

e e dus

π

π

π

� ��

� � �

� � �

�

�

�

� �

�

�Setting /2,x u s� � we get

2 2

2

/4

/2

/4

2[erf ( ); ]

1erfc ( /2)

s x

s

s

L t s e e dxs

e ss

π� ��

�

�(6.19)



EXAMPLE 6.14 Find the Laplace transform of erf (t1/2).

Solution From the definition of Laplace transform, we have

1/221/2

0 0

2[erf ( ); ]

� � ��tst uL t s e e du dt

π� �Changing the order of integration, we get

2

2

2 2

2

1/2

0

0

(1 )

0

2[erf ( ); ]

2

2

u st

u

u su

s u

L t s e du e dt

e dus

e dus

π

π

π

� ��

� � �

� � �

�

�

�

� �

�

�

Setting 2 2(1 ) or 1 ,s u x su x� � � � we have

/ 1du dx s� �

Then

2 21/2

0 0

2 1 2[erf ( ); ]

1 1x xL t s e dx e dx

s s s sπ π� ��

or

1/2 1[erf ( ); ]

1L t s

s s�

�(6.20)


cos t

t

Solution Let ( ) sin ;f t t� then

cos( )

2

tf t

t��

Using the property of the Laplace transform of the derivative of a function, we have

[ ( ); ] [ ( )] (0)L f t s sL f t f� �



Therefore,

cos; [sin ; ]

2

tL s sL t s

t

� ��

(6.21)

But,

3 5

3/2 5/21/2

3/2 5/2 7/2

3/2 5/2 7/2

3/2

( ) ( )[sin ; ] ;

3! 5!

;3! 5!

3/2 1 5/2 1 7/2

3! 5!

1/2 1 3/2 1/2 1 5/2 3/2 1/2

6 120

1 1 11

4 2! 42

t tL t s L t s

t tL t s

s s s

s s s

s ss

π π π

π

�

�

�

�

� � ��

� � ��

� ��

� ��

� ��

2 3

1/43/2

1 1

3! 4

2s

s

es

π

�

�

� � ��

� ��

Now substituting Eq. (6.22) into Eq. (6.21), we get

1/41/2

cos;

2 2st

L s et s

π ��

Therefore,

1/4cos; st

L s et s

π ��

Hence the result.

��

Bessel functions arise in several problems involving circular or cylindrical geometry. It istherefore useful to find the Laplace transform of Bessel functions of the first kind.




(i) 0 ( ),J t (ii) 0 ( ),tJ t (iii) 0 ( ).ate J t�

Solution (i) From the definition of the Bessel function, we have

2

0

( 1)( )

! 1 2

n rr

nr

tJ t

r n r

��

�

� � ��

For 0n � , we have

2 2

0 20 0

2 4 6

2 2 2 2 2 2

( 1) ( 1)( )

! 1 2 2( !)

12 2 4 2 4 6

r rr r

r r

t tJ t

r r r

t t t�

� �

� �

� ��

� � � � ��

Thus,

2 40 2 2 2

2 3 2 2 5 2 2 2 7

2 3

2 2 2

1/2

2 2

1 1[ ( ); ] [1; ] [ ; ] [ ; ]

2 2 4

1 1 2! 1 4! 1 6!

2 2 4 2 4 6

1 1 1 1 3 1 1 3 5 11

2 2 4 2 4 6

1 1 11

1

L J t s L s L t s L t s

s s s s

s s s s

s s s

�

�

�

�

� � � ��

� � � � ��

� � � ��

� ��

Hence,

0 2

1[ ( ); ]

1L J t s

s�

(ii) From the properties of Laplace transform, we have

[ ( ); ] ( 1) ( )n

n nn

dL t f t s F s

ds� �



Therefore,

0 0 2

1[ ( ); ] ( 1) [ [ ( ); ]]

1

d dL tJ t s L J t s

ds ds s

��

Thus,

0 2 3/2[ ( ); ]

(1 )

sL tJ t s

s�

(iii) From the shifting property of the Laplace transform, we have

[ ( ); ] ( )atL e f t s F s a� � �

Therefore,

0 2 2( )

1 1[ ( ); ]

1 1 ( )

at

s s a

L e J t ss s a

�

� �

� ��

��

The concept of impulse function or Dirac delta function has been introduced in Chapter 3itself. In certain applications involving a sudden excitation of a system or a large voltage overa short interval of time, the Laplace transform of Dirac Delta function is useful. From theproperty of Dirac Delta function, we have

0( ) ( ) ( )t a f t dt f a�

��

In particular, if ( ) ,stf t e�� then

0[ ( ); ] ( ) , 0st asL t a s e t a dt e a� �

� � �� (6.23)

��

So far we have discussed various properties of the Laplace transform and studied the Laplacetransform of some simple functions. However, if the Laplace transform technique is to beuseful in applications, we have to consider the reverse problem too, i.e., we have to find the

original function ( )f t when we know its Laplace transform ( ).F s Thus, if

[ ( ); ] ( )L f t s F s�

then

1( ) [ ( ); ]f t L F s t�� (6.24)



In other words, the inverse Laplace transform of a given function ( )F s is that function f (t)whose Laplace transform is ( ).F s It can be established that ( )f t is unique. Here, L–1 is knownas inverse Laplace transform operator. From the elementary definition (6.24) and from theresults obtained thus far in finding the Laplace transform of some elementary functions, wecan immediately generate the following table of transforms:

Table 6.2 Table of Laplace Transform

( )f t [ ( ); ] ( )L f t s F s� ( )F s 1[ ( ); ] ( )L F s t f t�

�

0 0 0 0

1 1/s 1/s 1

ate 1/( )s a− 1/( )s a− ate

ate� 1/( )s a+ 1/( )s a+ ate�

t 21/s 21/s t

nt 1!/ nn s + 11/ ns + / !nt n

sin at 2 2

a

s a� 2 2

a

s a�sin at

cos at 2 2

s

s a� 2 2

s

s a�cos at

sinh at 2 2

a

s a� 2 2

a

s a�sinh at

cosh at 2 2

s

s a� 2 2

s

s a�cosh at

sint at 2 2 2

2

( )

as

s a�2 2 2

2

( )

as

s a�sint at

cost at2 2

2 2 2( )

s a

s a

�

�

2 2

2 2 2( )

s a

s a

�

�

cost at

In most of the problems we have considered earlier, [ ( ); ]L f t s is a simple rational function.

The linearity property holds true even in the case of inverse transform. That is, if 1 2( ) and ( )F s F s

are the Laplace transform of 1 2( ) and ( ),f t t t and if c1 and c2 are any two constants, then

1 1 11 1 2 2 1 1 2 2[{ ( ) ( )}; ] [ ( ); ] [ ( ); ]L c F s c F s t c L F s t c L F s t� � �

� � �

By expressing [ ( ); ]L f t s as partial fractions, we should be able to recognise them as the

transform of some known functions, with the help of which we can write down the inversetransform. Similarly, shifting property is also useful in constructing the inverse transform ofsome functions, which is stated in the following theorem:



Theorem 6.10 If [ ( ); ] ( ),L f t s F s� then

1 1[ ( ); ] [ ( ); ]atL F s a t e L F s t� � �

� � (6.25)

Proof Since [ ( ); ] ( ),L f t s F s� we have

1[ ( ); ] ( )L F s t f t�

�

Recalling the shifting property of Laplace transform, we find that

1

[ ( ); ] ( )

[ ( ); ] ( )

at

at

L e f t s F s a

L F s a t e f t

�

� �

� �

� �

Thus,

1 1[ ( ); ] [ ( ); ]atL F s a t e L F s t− − −+ =

EXAMPLE 6.17 Obtain the inverse Laplace transform of

2

2

4 3 5

( 1)( 2 2)

s s

s s s

� �

� � �

Solution Using partial fraction expansion, we can write

2

2 2

4 3 5

1( 1)( 2 2) 2 2

s s A Bs C

ss s s s s

� � ��

��

Therefore,

2 24 3 5 ( 2 2) ( ) ( 1)s s A s s Bs C s� � � � � � � �

Let s = –1; then A = 12/5. Equating the coefficient of s on both sides, we have

9/5B C+ =

Equating the coefficient of constant on both sides, we get 2 5A C� � which gives 1/5,C =

and hence 8/5.B = The given expression can now be written as

2

2 2 2 2 2

4 3 5 12 1 8 1 1

5 1 5 5( 1)( 2 2) ( 1) 1 ( 1) 1

s s s

ss s s s s

� ��

��



Thus, we find that

21 1 1 1

2 2 2 2 2

1 12

12

4 3 5 12 1 8 ( 1) 1 1 1; ; ; ;

5 1 5 5( 1) ( 2 2) ( 1) 1 ( 1) 1

12 1 8 1; ;

5 5 1

1 1; (by using the shifting property)

5 1

t t

t

s s sL t L t L t L t

ss s s s s

se L t e L t

s s

e L ts

− − − −

− − −

−

⎡ ⎤− + − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦+ − + − + − +⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

+⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎣ ⎦ +⎣ ⎦

⎡ ⎤+ ⎢ ⎥+⎣ ⎦

12 8 1(cos sin ) sin

5 5 5

12 8 9cos sin

5 5 5

t t t

t t t

e e t t e t

e e t e t

�

�

� � � �

� � �

Theorem 6.11 If ( )f t is a piecewise continuous function and satisfies the condition of

exponential order 0c such that0

Lt ( )/t

f t t→

exists, then for 0 ,s c�

1

0

( ); ( )

tF sL t f x dx

s− ⎡ ⎤ =⎢ ⎥⎣ ⎦ ∫

Proof Let0

( ) ( ) .t

G t f x dx= ∫ Then (0) 0 and ( ) ( );G G t f t�� Also,

[ ( ); ] [ ( ); ] (0) [ ( ); ]L G t s sL G t s G sL G t s� � � �

i.e.,

( ) [ ( ); ]F s sL G t s�

Therefore,

( )[ ( ); ]

F sL G t s

s�

Hence,

1

0

( ); ( ) ( )

tF sL t G t f x dx

s− ⎡ ⎤ = =⎢ ⎥⎣ ⎦ ∫ (6.26)

This result can be generalized to show that

1

0 0 0 0

( ); ( )

t t t t nn

F sL t f t dt

s− ⎡ ⎤ =⎢ ⎥⎣ ⎦ ∫ ∫ ∫ ∫� (6.27)



Theorem 6.12 (Change of scale property). If

1[ ( ), ] ( )L F s t f t− =

then

1 1[ ( ); ]

tL F s t f�

� �

�

� ��


0( ) ( )stF s e f t dt�

� ��

Therefore,

0( ) ( )stF s e f t dt��

�

� ��

Let ,t x� � so that / .dt dxα= Then we get

0

1 1 1( ) ; ;sx x x t

F s e f dx L f s L f s�

� � � � � �

�

� � � ��

Thus,

1 1[ ( ); ]

tL F s t f�

� �

�

� ��

(6.28)

EXAMPLE 6.18 Find the inverse Laplace transform of

(i)1

,( )ns a�

(ii)2

2,

4 13

s

s s

�

� �

(iii)2

2 3,

3/4

s

s s

−− −

(iv)3

2 2 2.

( )

s

s a�

Solution (i) Using the shifting property, we at once have

11 11 1

; ;( 1)!( )

at nat

n n

e tL t e L t

ns a s

� �

� � �

� � � ��

(ii)2 2 2

2 ( 2) 4

4 13 ( 2) 3

s s

s s s

� � ��

� � � �

Therefore,

1 1 12 2 2 2 2

2 1 2 12 2 2 2

2 ( 2) 1; ; 4 ;

4 13 ( 2) 3 ( 2) 3

1 (by using the; 4 ;

shifting property)3 3t t

s sL t L t L t

s s s s

se L t e L t

s s

� � �

� �

� ��

� � � � � ��

� � � ��

� ��



Thus,

1 2 22

2; cos 3 (4/3) sin 3

4 13t ts

L s e t e ts s

− +⎡ ⎤ = +⎢ ⎥− +⎣ ⎦

(iii)2 2 2

2 3 2 3 2( 1/2) 2

3/4 ( 1/2) 1 ( 1/2) 1

s s s

s s s s

− − − −= =− − − − − −

Thus,

1 1 12 2 2

/2 1 /2 12 2 2 2

2 3 ( 1/2) 1; 2 ; 2 ;

3/4 ( 1/2) 1 ( 1/2) 1

1 (by using the2 ; 2 ;

shifting property)1 1t t

s sL t L t L t

s s s s

se L t e L t

s s

− − −

− −

− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

Therefore,

1 /2 /22

2 3; 2 cosh 2 sinh

3/4t ts

L t e t e ts s

− −⎡ ⎤ = −⎢ ⎥− −⎣ ⎦

(iv)3 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2

( )

( ) ( ) ( )

s s s a a s a s

s a s a s a s a

� ��

� � � �

Hence,

31 1 2 1

2 2 2 2 2 2 2 2; ; ;

( ) ( )

s s sL t L t a L t

s a s a s a− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ + +⎣ ⎦⎢ ⎥⎣ ⎦ ⎣ ⎦

Thereore,

31

2 2 2; cos sin

2( )

s aL t at t at

s a�

� ��

��


(i)2 1 ,ln

( 1)

s

s s

++

(ii) 1cot .s

k− ⎛ ⎞⎜ ⎟⎝ ⎠

Solution (i) From Theorem 6.4, we have

( )[ ( ); ] ( 1) ( )n n nL t f t s F s� �



In particular, 1n � gives

[ ( ); ] ( )L tf t s F s��

i.e.

( ) [ ( ); ]d

F s L tf t sds

� � (6.29)

Let

2 1[ ( ); ] ln ( )

( 1)

sL f t s F s

s s

+= =+

Then,

2

2

( ) [ln ( 1) ln ln ( 1)]

2 1 1[ ( ); ]

11

d dF s s s s

ds ds

sL tf t s

s ss

� � � � �

� � � � �

��

Now, using Eq. (6.29), we get

2

1 1 2[ ( ); ]

1 1

sL tf t s

s s s� � �

� �

Hence,

1 1 12

1 1( ) ; ; 2 ;

1 1

1 2 cost

stf t L t L t L t

s s s

e t

− − −

−

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ +⎝ ⎠

= + −

Therefore,

21 1 1 2 cos

( ) ln ;( 1)

ts e tf t L t

s s t

�

�

� ��

��

(ii) Let 1[ ( ); ] cot ( )s

L f t s F sk

�

� ��

Then,

12 2

( ) cotd d s k

F sds ds k k s

�

� ��



Now using Eq. (6.29), we obtain

2 2( ) [ ( ); ]

d kF s L tf t s

ds k s= =

+

Therefore,

12 2

( ) ; sink

tf t L t ktk s

�

� ��

��

Hence,

1 1 sin( ) cot ;

s ktf t L t

k t− −⎡ ⎤⎛ ⎞= =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

�� !��"�� #

We often come across functions which are not the transform of some known function, butthen, they can possibly be expressed as a product of two functions, each of which is the

transform of a known function. Thus we may be able to write the given function as ( ) ( ),F s G s

where ( ) and ( )F s G s are known to be transforms of the functions ( ) and ( ),f t g t respectively.

Theorem 6.13 If ( ) and ( )F s G s are the Laplace transforms of ( ) and ( )f t g t respectively,

then ( ) ( )F s G s is the Laplace transform of

0( ) ( )

tf t u g u du−∫

i.e.

1

0[ ( ) ( )] ( ) ( )

tL F s G s f t u g u du− = −∫ (6.30)

This integral is called the convolution of andf g and is denoted by the symbol * .f g


0 0

( )

0 0

( )

0 0

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

sv su

s v u

v u

F s G s e f v dv e g u du

e f v g u dv du

g u e f v dv du

� �

� �

� �

� � � ��

�

� � �

�

� �

� �

� �

� �

� �

� �



Let u v t� � in the inner integral. Then,

0( ) ( ) ( ) ( )st

uF s G s g u e f t u dt du�

� ��

� ��

Change the order of integration as shown in Fig. 6.3.

t

ut

=

u

O

t u = t = �

Fig. 6.3 Convolution integral.

Then, we get

0 0

0 0

0

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ;

t st

tst

t

F s G s e f t u g u du dt

e f t u g u du dt

L f t u g u du t

�

�

� ��

� �

� ��

� �

� ��

� �

� �

�

�

�

Hence the result.

Definition 6.3 We define the Laplace convolution of ( ) and ( )f t g t by the integral

0* ( ) ( )

tf g f t u g u du= −∫ (6.31)

It can be verified that f and g can be interchanged in the convolution, i.e., f and g are

commutative. Let t u v� � in Eq. (6.31) so that .du dv� � Then,

0

0* ( ) ( ) ( ) ( )

t

tf g f v g t v dv g t v f v dv= − − = −∫ ∫

Therefore,

* *f g g f= (6.32)



EXAMPLE 6.20 Apply the convolution theorem to evaluate

(i) 12 2 2

;( )

sL t

s a�

� ��

�� (ii) 1

2;

( ) ( 1)

sL t

s a s− ⎡ ⎤

⎢ ⎥+ +⎣ ⎦(iii) 1

2

1; .

( 1)L t

s s�

� ��

��

Solution Consider

(i)2 2 2 2 2 2 2

1

( )

s s a

as a s a s a

� ��

� � ��

Now choosing

2 2 2 2, ,( ) ( )

s aF s G s

s a s a= =

+ +

we can write the inverse transform for ( ) and ( )F s G s as

12 2

; cos ( )s

L t at f ts a

�

� � � ��

12 2

; sin ( )a

L t at g ts a

�

� � � ��

Hence, using the convolution theorem, we obtain

12 2 2 0

0

0 0

0 0

2

1; cos ( ) sin

( )

1(cos cos sin sin ) sin

1 sincos sin 2 (1 cos 2 )

2 2

cos cos 2 sin sin 2

2 2 2 2

1(cos cos 2 sin sin 2

4

t

t

t t

t t

sL t a t u au du

as a

at au at au au dua

atat au du au du

a a

at au at auu

a a a a

at at ata

− ⎡ ⎤ = −⎢ ⎥+⎣ ⎦

= +

= + −

⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − +

∫

∫

∫ ∫

2

cos sin)

24

sin

2

at t atat

aa

t at

a

+ +

=

Therefore,

12 2 2

sin;

2( )

s t atL t

as a�

� ��

��



(ii) Consider

2 2

1

( ) ( 1) 1

s s

s as a s s�

��

Now choosing

1 ,( )F ss a

=+ 2

( )1

sG s

s=

+the inverse transforms for this pair are obtained as

1

12

1; ( )

; cos ( )1

atL t e f ts a

sL t at g t

s

− −

−

⎡ ⎤ = =⎢ ⎥+⎣ ⎦

⎡ ⎤ = =⎢ ⎥+⎣ ⎦


1 ( )2 0

0

20

1; cos

( ) ( 1)

cos (a standard integral)

(sin cos )1

t a t u

tat au

tauat

L t e u dus a s

e e u du

ee u a u

a

− − −

−

−

⎡ ⎤ =⎢ ⎥+ +⎣ ⎦

=

⎡ ⎤= +⎢ ⎥

+⎣ ⎦

∫

∫

Therefore,

12 2

1; ( cos sin )

( ) ( 1) 1ats

L t a t t aes a s a

� �

� ��

� � ��

(iii) We shall write

2 2 2 2

1 1 1

( 1) ( 1)s s s s�

� �

For this pair, we can write inverse transforms as

12

1 12 2

1; ( )

1 1; ; ( )

( 1)t t

L t t f ts

L t e L t e t g ts s

−

− − − −

⎡ ⎤ = =⎢ ⎥⎣ ⎦

⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥+ ⎣ ⎦⎣ ⎦




12 2 0

2

0 0

20 00 0

1; ( )

( 1)

( ) [ ] 2

( 2) 2

t u

t tu u

t tt tu u u u

t

L t t u ue dus s

t e u du u e du

t ue e du u e ue du

t e t

− −

− −

− − − −

−

⎡ ⎤ = −⎢ ⎥+⎣ ⎦

= −

⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

= + + −

∫

∫ ∫

∫ ∫

Therefore

12 2

1; ( 2) 2

( 1)tL t t e t

s s� �

� ��

��

EXAMPLE 6.21 Prove that

0 00

( ) ( ) sintJ t J t u du t− =∫

Solution From Table 6.2, we note that

12

1; sin

1L t t

s�

� ��

��

We shall also write

2 2 2

1 1 1

1 1 1s s s�

� � �

Now taking

2 2

1 1( ) , ( )

1 1F s G s

s s� �

� �

their inverse transforms give

102

1; ( ) ( ) ( )

1L t J t f t g t

s

�

� ��

��

Hence, using the convolution theorem, we have

10 02 0

1; ( ) ( ) sin

1

tL t J t J t u du t

s− ⎡ ⎤ = − =⎢ ⎥+⎣ ⎦ ∫



��$ ��

Definition 6.4 The unit step function or Heaviside unit function is defined as

0 for( )

1 for , 0

t aH t a

t a a

��

� ��

Graphically, it can be depicted as in Fig. 6.4.

t

H t( – )a

1

aO

Fig. 6.4 Illustration of Heaviside unit function.

EXAMPLE 6.22 Find the Laplace transform of unit step function.

Solution From the definition of Laplace transform, we have

0

0

[ ( ); ] ( )

0 1

, 0

st

a st st

a

as

L H t a s e H t a dt

e dt e dt

es

s

−

− −

−

− = −

= ⋅ ⋅ +

= >

∫

∫ ∫

�

�

Theorem 6.14 (Second shifting property). If [ ( ); ] ( ),L f t s F s� then

[ ( ) ( ); ] ( )asL f t a H t a s e F s�

� � �

1[ ( ); ] ( ) ( )asL e F s t f t a H t a� �

� � �


0

0

[ ( ) ( ); ] ( ) ( )

0 ( )

st

a st

a

L f t a H t a s e f t a H t a dt

dt e f t a dt

�

�

� � � � �

� � � �

�

� �

�

�



Let t – a = v, and hence dt = dv, the right-hand side of the above equation becomes

( )

0 0( ) ( )

( )

s a v as sv

as

e f v dv e e f v dv

e F s

� ��

�

�

�

� �

Therefore,

[ ( ) ( ); ] ( )asL f t a H t a s e F s�

� � � (6.33)

1[ ( ); ] ( ) ( )asL e F s t f t a H t a� �

� � � (6.34)


2, 0

1

asea

s

�

�

�

Solution From the second shifting property, we have

1[ ( ); ] ( ) ( )asL e F s t f t a H t a� �

� � �

In the given problem,

2

1( )

1F s

s�

�

Hence,

1 12

1( ) [ ( ); ] ; sin

1f t L F s t L t t

s� �

� ��

��

Therefore,

12

1sin ( ) ( )

1

0,

sin ( ),

asL e t a H t as

t a

t a t a

� �

� ��

��

��

� �

Theorem 6.15 (Heaviside expansion theorem). Let ( ) and ( )F s G s be two polynomials in s

where the degree of ( )F s is lower than that of ( )G s and if G(s) has n distinct

roots ( 1, , ),i i nα = … then



1

1

( )( );

( ) ( )i

nti

ii

FF sL t e

G s Gαα

α−

=

⎡ ⎤ =⎢ ⎥ ′⎣ ⎦∑ (6.35)

Proof Since ( )G s is a polynomial in s having n distinct roots and F(s) is a polynomial

whose degree is less than that of ( )G s , we have

1 2

1 2

1

( ) ( )

( )( )

nn

ni

i

AA AF s F s

G s s s ss

α α αα=

= = + + +− − −Π −

� (using partial fractions)

Multiplying both sides by ( )is �� and taking the limit as ,is �� we obtain the coefficients

( ) ( )Lt ( ) Lt

( ) ( )i i

i ii i

s s

F s s sA F

G s G s� �

� �

�

� �

� �

� �

which take indeterminate form and, therefore, using L’Hospital’s rule, we get

( )1( ) Lt

( ) ( )i

ii i

s i

FA F

G s G�

�

�

��

� �

� �

Hence,

1 2

1 1 2 2

( )( ) ( )( ) 1 1 1

( ) ( ) ( ) ( ) ( ) ( ) ( )n

n n

FF FF s

G s G s G s G s

αα αα α α α α α

= + + +′ ′ ′− − −

�

Thus,

1 1 11 2

1 1 2 2

1

1

( ) ( )( ) 1 1; ; ;

( ) ( ) ( )

( ) ( )1;

( ) ( )

�

in

tn i

n n ii

F FF sL t L t L t

G s G s G s

F FL t e

G s Gα

α αα α α α

α αα α α

− − −

−

=

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥′ ′− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤+ =⎢ ⎥′ ′−⎣ ⎦∑ (6.36)

Hence the result.

EXAMPLE 6.24 Using the method of Heaviside expansion theorem, find the Laplace inverseof

2

3 2

1

3 2

s

s s s

++ +

Solution We shall take 2( ) 1,F s s� � and

3 2( ) 3 2 ( 1) ( 2)G s s s s s s s� � � � � �



Here, ( )G s has three distinct roots 0, 1, 2� � and the degree of ( )F s is lower than that of

( ).G s Hence, using the Heaviside expansion theorem, we have

21 0 2

3 2

1 (0) ( 1) ( 2);

(0) ( 1) ( 2)3 2r t ts F F F

L t e e eG G Gs s s

� � �

� ��

� � ��

But, 2( ) 3 6 2.G s s s′ = + + Therefore,

21 2

3 2

1 1 5; 2

2 23 2t ts

L t e es s s

� � �

� ��

� ��

��

In solving some complicated problems using the Laplace transform method, the direct approachso far followed may not be helpful in finding the inverse Laplace transform. Methods basedon complex variable theory may come in handy for finding the inverse transform. Also, it can

be noted that the Laplace transform of ( )f t is expressed as an integral. Similarly, the inverse

Laplace transform of ( )F s can be expressed as in integral which is known as inverse integral.

This integral can be evaluated by using contour integration methods. The complex inversionformula is stated in the following theorem.

Theorem 6.16 Let ( ) and ( )f t f t� be continuous functions on 0t � and ( ) 0 for 0.f t t= < In

addition, if 0( ) is 0( )tf t eγ and

( ) [ ( ); ]F s L f t s=

Then

1 1[ ( ); ] ( ) ( )

2

i st

iL F s t f t e F s ds

i

�

��

��

�

� � ��

�

0 andt γ> is a positive constant.

Proof Let ( )g t and ( )g t� be continuous functions and if ( )g t dt�

�� converges absolutely

and uniformly then ( )g t may be represented by the Fourier integral formula

1( ) ( ) cos ( )

2

1( ) cos ( )

2

g t g t d d

g t d d

ν ω ν ω νπ

ν ω ν ν ωπ

− −

− −

⎡ ⎤= −⎢ ⎥⎣ ⎦

⎡ ⎤= −⎢ ⎥⎣ ⎦

∫ ∫

∫ ∫

� �

� �

� �

� �(6.37)



Since sin ( )tω ν− is an odd function of ,� we have

1( ) sin ( ) 0

2g t d d� � � � �

� � �

� ��

� �

� �

Combining this expression with Eq. (6.37), we can write

( )1( ) ( )

2

1( )

2

i t

i t i

g t g e d d

e g e d d

� �

� ��

� � �

�

� � �

�

�

� �

�

� �

� ��

� ��

� �

� �

� �

� �

� �

� �(6.38)

In addition, we assume that ( )g t is of exponential order 00( ).te��

Now we consider the function

( ), 0( )

0, 0

te f t tg t

t

γ−⎧ ≥⎪= ⎨<⎪⎩

where � is a real number greater than 0.� Thus, ( )g t satisfies all the conditions required by

the Fourier integral theorem and, therefore, we have from Eq. (6.38), for 0t � the relation

( )

1( ) ( )

2

1( )

2

1( ) [From the definition of Laplace transform]

2

t i t i

i t i

i t

e f t e e f e d d

e e f d d

e F i d

� � ��

� � � �

�

� � ��

� � ��

� � ��

� �

� �

� �

� �

�

� ��

� ��

� �

� �

� �

�

� �

� �

� �

� �

�

�

Let ,i s� �� so that / .d ds iω = It follows that

( )1( ) ( )

2

it t s

ie f t e F s ds

i

��

��

��

�

� ��

�

Therefore,

1( ) ( ) , 0

2

i st

if t e F s ds t

i

�

��

�

�

� ��

�(6.39)

Hence the proof.




2

1

( 1) ( 2)s s+ −

Solution From complex inversion formula, we get

12 2

2

1 1;

2( 1) ( 2) ( 1) ( 2)

sum of the residues of( 1) ( 2)

sti

i

st

e dsL t

is s s s

e

s s

γ

γπ+−−

⎡ ⎤ =⎢ ⎥+ − + −⎣ ⎦

=+ −

∫�

�

at the simple pole 1s � � and double pole 2.s � Therefore,

12 21 2

22

2 2

1 ( 1); Lt Lt

1( 1) ( 2) ( 1) ( 2)

{( 1) 1}Lt

9 ( 1)

1

9 3 9

st st

s s

t st

s

tt t

s e d eL t

ds ss s s s

e e s t

s

e te e

−→− →

−

→

−

⎛ ⎞+⎡ ⎤ = + ⎜ ⎟⎢ ⎥ +⎝ ⎠+ − + −⎣ ⎦

⎡ ⎤+ −= + ⎢ ⎥+⎢ ⎥⎣ ⎦

= + −


sinh ( ), 0

sinh ( )

x sx l

l s� �

Solution Let

sinh ( )( )

sinh ( )

x sF s

l s�

Then,

1 1 sinh ( ) 1 sinh ( )[ ( ); ]

2 sinh ( ) 2 sinh ( )

i st st

i C

x s x sL F s t e ds e ds

i l s i l s

�

��

��

�

� ��

�

= sum of the residues of the integrand at infinitely many simple

poles given by the roots of transcendental equation sinh ( ) 0l s �



i.e.,

20, implying 2 1 .l s l s n ie e l s e ��

� � � �

Hence, 2 2 2 ,sl n �� i.e.,

2 2

2, 0,1, 2, ...

ns n

l

��

Thus, we have to compute the residues at the poles

0,s �2 2

2,

nn

s sl

π�� 1, 2,n � �

Now, the residue at 0s � is

0

sinh ( )Lt 0

sinh ( )st

s

x sse

l s�

�

The residues at ns s� are obtained as

2 2

2 2 2 22

2 22 2

2

2 2

2 2

2 2

2 2

sinh ( ) 2Lt ( ) sinh ( )

sinh ( ) cosh ( )

2exp sinh

cosh

2 sinh [ ( / ) ]exp

cosh ( )

2 ( 1) sexp

nn

st stn s ss s

x s ss s e e x s

l s l l s

nn t nl xl ln

l ll

in n t i n l x

inl l

n n t

l l

ππ π

π

π π ππ

π π

=→− =

−⎛ ⎞⎛ ⎞− −⎜ ⎟= ⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠ ⎝ ⎠−⎜ ⎟⎜ ⎟⎝ ⎠

⎛ ⎞−= ⎜ ⎟⎝ ⎠

⎛ ⎞− −= ⎜ ⎟⎝ ⎠

1 2 2

2 2

in ( / )

cos

( 1) 2sin exp ; 1, 2, ...

n

n x l

n

n n n tx n

ll l

ππ

π π π+ ⎛ ⎞− ⋅ −⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Therefore,

2 21 1

2 21

sinh ( ) 2; ( 1) sin exp

sinh ( )n

n

x s n n tL t n x

l s ll l

� � ��

�

� ��



��

The Laplace transform technique is one of the powerful tools for solving physical problemsinvolving ordinary differential equations (ODE), particularly initial value problems. It reducesthe solution of ODE to the solution of an algebraic equation. This method has a particularadvantage in finding the solution of an initial value problem, without first finding the generalsolution and then using the given ICs for evaluating the arbitrary constants. We shall firstapply the Laplace transform technique to find the solution of a typical initial value problemand demonstrate the steps involved. Consider a second order linear differential equation

2

2( )

d y dyy f t

dtdt� �� (6.40)

subject to the ICs

(0) , (0)y A y B�� (6.41)

where and� � are constants. Taking the Laplace transform of (6.40) on both sides, we obtain

[{ }; ] [ ( ); ]L y y y s L f t s� ��

Using the property of the Laplace transform of the derivatives, we get

2{ [ ( ); ] (0) (0)} { [ ( ); ] (0)} [ ( ); ] [ ( ); ]s L y t s sy y sL y t s y L y t s L f t s� ��

Using the ICs, we have, after regrouping the terms,

2( ) [ ( ); ] [ ( ); ] ( )s s L y t s L f t s s A B� � ��

Alternatively,

2( ) ( ) ( ) ( )s s Y s F s s A B� � ��

i.e.

2 2

( ) ( )( )

s A B F sY s

s s s s

�

� � � �

� ��

� � � �

(6.42)

Taking the inverse Laplace transform on both sides, we get the solution in the form

1 12 2

( ) ( )( ) ; ;

s A B F sy t L t L t

s s s s

�

� � � �

� �

� ��

� � � �� (6.43)

EXAMPLE 6.27 Solve the equation using the Laplace transform method

4 8 cos 2y y y t��

Given that 2 and 1 when 0.y y t��



Solution Taking the Laplace transform of the given ODE, we obtain

22

{ ( ) (0) (0)} 4{ ( ) (0)} 8 ( )4

ss Y s sy y sY s y Y s

s��

�

Using the initial conditions, we get

22

( ) 2 1 4 ( ) 8 8 ( )4

ss Y s s sY s Y s

s� � � � � �

�

22

( 4 8) ( ) 2 94

ss s Y s s

s� � � � �

�

Therefore,

2 2 2 2

2 2 2 2

2 2 2 2 2 2

( /20 1/5) ( /20 2/5) 2 9( )

4 4 8 4 8 4 8

1 1 1 1 ( 2) 2 2

20 5 20 54 4 ( 2) 2

1 2( 2) 4 9

( 2) 2 ( 2) 2 ( 2) 2

s s sY s

s s s s s s s

s s

s s s

s

s s s

+ += − + ++ + + + + + +

+ −= × + − × −+ + + +

+ −× + ++ + + + + +

Taking the inverse Laplace transform, we obtain

2 2

2 2 2 2

1 1 1 1( ) cos 2 sin 2 cos 2 sin 2

20 10 20 20

2 9sin 2 2 cos 2 2 sin 2 sin 2

10 2

t t

t t t t

y t t t e t e t

e t e t e t e t

− −

− − − −

= + − +

− + − +

On simplification, we get

2 1( ) (39 cos 2 47 sin 2 ) (cos 2 2 sin 2 )

20 20

tey t t t t t

−= + + +

EXAMPLE 6.28 Using the Laplace transform technique, solve

3 2 tx x x te��

Given

(0) 1 and (0) 0x x� ��

Solution Taking the Laplace transform of the given ODE, we get

22

1{ ( ) (0) (0)} 3{ ( ) (0)} 2 ( )

( 1)s X s sx x sX s x X s

s��

�



Applying the ICs, we have

22

1( 3 2) ( ) 3

( 1)s s X s s

s� � � � �

�

Therefore,

3

2 3

1 3( )

( 1) ( 2)( 1) ( 2)

1 1 1 1 1 1

2 1 2 2( 1) ( 1)

1 1 2 1(using partial fractions)

2 2 1 2

sX s

s ss s

s s s

s s s

+= ++ ++ +

= × − × + ×+ + +

− × + −+ + +

The inverse Laplace transform yields

12 3

2 22 2

5 1 1 1 1 1 3 1( ) ;

2 1 2 2 2 2( 1) ( 1)

5 1 1 3 35

2 2 2 2 2 2 2 2

tt t t t t

x t L ts ss s

t e te e t e e t e

�

�

� � � � �

� ��

� ��

� �

EXAMPLE 6.29 Using the Laplace transform technique, solve the following initial valueproblem:

0,ty y ty′′ ′+ + = (0) 1,y = (0) 0y′ =

Solution This is an example of ODE with variable coefficients. Taking the Laplacetransform on both sides of the given ODE and using Eq. (6.7), we get

[ ; ] [ ; ] [ ; ] 0d d

L y s L y s L y sds ds

��

2{ ( ) (0) (0)} { ( ) (0)} { ( )} 0d d

s Y s sy y sY s y Y sds ds

��

Applying the ICs, we obtain

2{ ( ) } { ( ) 1} ( ) 0d d

s Y s s sY s Y sds ds

� � � � �

or

2( 1) 0dY

s sYds

� � �



which is a first order ODE. On rewriting, we get

2 01

dY s dsY s

+ =+

Integrating, we see that

21ln ln ( 1) ln2

Y s c+ + =

2 1

cYs

=+

Now, taking the inverse Laplace transform, we find 0( ) ( ),y t cJ t= where 0 ( )J t is a Bessel

function of order zero. Since 0(0) 1 (0) ,y cJ c= = = the required solution is

0( ) ( )y t J t=

EXAMPLE 6.30 Solve the simultaneous equations

, sintdx dyy e x tdt dt

− = + =

using the Laplace transform method which satisfies the conditions (0) 1, (0) 0.x y= =

Solution Taking the Laplace transform of both the equations and using the notation

[ ; ],Y L y s= we have

2

1(0) [ ; ]1

1(0) [sin ; ]1

tsX x Y L e ss

sY y X L t ss

− − = =−

− + = =+

Using the given ICs, the above equations reduce to

1ssX Y

s− =

− (6.44)

21

1X sY

s+ =

+(6.45)

Solving Eqs. (6.44) and (6.45), we get

2

2 2 21

( 1) ( 1) ( 1)sX

s s s= +

− + +(6.46)



3 2

2 2

2

( 1) ( 1)

s s sY

s s

� � ��

� �

(6.47)

2 2 2( 1) ( 1) ( 1)

s s

s s s= −

+ − +(6.48)

Using partial fractions, we get2

2 2 2 2

1 1 1

( 1) 2 1( 1) ( 1) 1 1 1

s A Bs C s

s ss s s s s

� � �� Hence Eq. (6.46) gives

2 2 2 2

1 1 1 1

2 1 1 1 ( 1)

sX

s s s s

� ��

Taking inverse Laplace transform, we obtain

1[ cos sin (sin cos )]

2tx e t t t t t� � � � � (6.49)

Also, from Eq. (6.48), we have

2 2 2 2

1 1 1

2 1( 1) 1 1

s sY

ss s s

� ��

Again, taking inverse Laplace transform, we get

1( sin cos sin )

2ty t t e t t� � � � (6.50)

Equations (6.49) and (6.50) constitute the solution of the given system.

��

A large number of problems in science and engineering involve the solution of linear partialdifferential equations. A function of two or more variables may also have a Laplace transform.Suppose x and t are two independent variables; consider t as the principal variable and x asthe secondary variable. When the Laplace transform is applied with t as a variable, the PDE

is reduced to an ordinary differential equation of the t-transform ( , ),U x s where x is the

independent variable. The general solution ( , )U x s of the ODE is then fitted to the BCs of

the original problem. Finally, the solution ( , )u x t is obtained by using the complex inversion

formula. Thus, the Laplace transform is specially suited to solving initial boundary value

problems (IBVP), when conditions are prescribed at 0.t � We have already noted in Chapters 3and 4 that such situations naturally arise in the case of heat conduction equation and wave



equation. Before we consider the Laplace transform method of solution of IBVP, we considerthe following example to prove some of the useful elementary results.

EXAMPLE 6.31 If ( , )u x t is a function of two variables x and t, prove that

(i) ; ( , ) ( , 0)u

L s sU x s u xt

� � � ��

(ii)2

22

; ( , ) ( , 0) ( , 0)tu

L s s U x s su x u xt

� ��

� ��

(iii) ( , );

u dU x sL s

x dx⎡ ⎤ =⎢ ⎥⎣ ⎦��

(iv)2 2

2 2; ( , )

u dL s U x s

x dx

� ��

��

(v)2

; ( , ) ( , 0)u d d

L s s U x s u xx t dx dx

��

� �

��

.

where ( , ) [ ( , ); ].U x s L u x t s�

Proof Following the definition of the Laplace transform, we have

(i)0 0

0 0

0

; Lt

Lt { ( , )} ( , )

( , 0) ( , )

pst st

p

ppst st

p

st

u u uL s e dt e dt

t t t

e u x t s e u x t dt

u x s e u x t dt

� � ��

� ��

� �

� � � ��

� ��

� � �

� �

�

�

� � ��

Therefore,

; ( , ) ( , 0)u

L s sU x s u xt

� � � ��

(ii)2

2; ; ,

[ ; ] ( , 0)

{ ( , ) ( , 0)} ( , 0)t

u V uL s L s V

t tt

sL V s V x

s sU x s u x u x

� � � ��

� � �

� � ��



Thus,

22

2 ; ( , ) ( , 0) ( , 0)tuL s s U x s su x u x

t

⎡ ⎤= − −⎢ ⎥

⎣ ⎦∂∂

(iii)0 0

; ( , ) ( , )st stu u d dUL s e dt e u x t dt x sx x dx dx

∞ ∞− −⎡ ⎤ = = =⎢ ⎥⎣ ⎦ ∫ ∫∂ ∂∂ ∂

(iv)2 2

2 2; ; ( , ),u u d dU d U uL s L s x s ux dx dx xx dx

⎡ ⎤ ⎡ ⎤ ⎛ ⎞= = = =⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦⎣ ⎦∂ ∂ ∂

∂ ∂∂

(v)2

; [ ( , ) ( , 0)] ( , ) ( , 0)u d dU duL s sU x s u x s x s xx t dx dx dx

⎡ ⎤= − = −⎢ ⎥

⎣ ⎦

∂∂ ∂

Thus, we notice from the above results that the partial derivatives are transformed into ordinaryderivatives.

6.13.1 Solution of Diffusion Equation

EXAMPLE 6.32 Solve the following IBVP using the Laplace transform technique:

PDE: , 0 1, 0t xxu u x t= < < >

BCs: (0, ) 1, (1, ) 1, 0u t u t t= = >

, 0 1IC: ( ) 1 sin , 0u x x xπ <= + <

Solution Taking the Laplace transform of both sides of the given PDE, we have

2

2( , ) ( , 0) d UsU x s u xdx

− =

Thus, the solution of the second order PDE reduces to the solution of second order ODE givenby

2

2 ( , ) 1(1 sin ) (after using IC)d U sU x s xdx

π− = − + (6.51)

The general solution of Eq. (6.51) is found to be

21 sin( , ) sx sx xU x s Ae Bes s

ππ

−= + + ++

(6.52)

But, (0, ) 1, (1, ) 1,u t u t= = and their Laplace transforms are

1 1(0, ) , (1, )U s U ss s

= =



From Eq. (6.52), we have

1 1A B

s s� � �

Hence, 0,A B� � and

1 1s sAe Bes s

�

� � �

Therefore,

0s sAe Be��

This is a homogeneous system; the determinant of the coefficient matrix is

1 10s s

s se e

e e

�

�

� � �

Thus, the only possible solution is the trivial solution and, therefore,

0A B� �

From Eq. (6.52), we now have

2

1 sin( , )

xU x s

s s

�

�

� �

�

(6.53)

Taking the inverse Laplace transform of Eq. (6.53), we get

1 12

1 sin( , ) ; ;

xu x t L t L t

s s

�

�

� �

� � � ��

Thus,

2( , ) 1 sin tu x t x e ππ −= + (6.54)

is the required solution.

EXAMPLE 6.33 Using Laplace transform, solve the following initial boundary value problem:

PDE: ,t xxku u= 0 1,x� � 0 t� � �

BCs: (0, ) 0,u t � ( , ) ( ),u l t g t� 0 t� � �

IC: ( , 0) 0,u x = 0 x l< <

Solution Taking the Laplace transform of both sides of the PDE, we get

2

2[ ( , ) ( , 0)]

d UK sU x s u x

dx− =



Thus the solution of PDE reduces to the solution of ODE which, after using the IC, can berewritten as

2

2( , ) 0

d UKsU x s

dx− = (6.55)


cosh ( ) sinh ( )U ks x B ks x= + (6.56)

Taking the Laplace transform of the BCs, we have

(0, ) 0, ( , ) ( )U s U l s G s� � (6.57)

Applying Eq. (6.57) in Eq. (6.56), we get 0 = A. Therefore,

( ) sinh ( )G s B ks l�

which implies

( )

sinh ( )

G sB

ks l� (6.58)


1 sinh ( )( , ) ( )

2 sinh( )

i st

i

ks xu x t G s e ds

i ks l

�

��

�

�

� ��

�(6.59)

To evaluate the integral on the right-hand side of this equation, we use the method of residuesfor which we note that the integrand has poles given by

2 2

sinh ( ) 0

1

ks l ks l

ks l n i

ks l e e

e e π

−= = −

= =

implying the simple poles at

2 2

2, 0, 1, 2, .n

ns n

Kl

π= − = ± ± … (6.60)

Now,

1 sinh ( )( )

2 sinh ( )st ks x

G s e dsi ks l

�

��

�

�

��

�sum of the residues of the integrand at the poles.



But the residue at the pole 0s � is zero. The residues at ns s� are

2 2 2 2

2 2 2 22 2

2 22 2

2

2 2

2

2

sinh ( )Lt ( )

[cosh ( )]

2 ( )Lt sinh ( )

cosh ( )

2exp sinh , 1, 2, 3, ...

cosh

2sinh

cosh ( )

n

n

st

s s

st

s s

ks xG s e

dks l

ds

sG se ks x

l k ks l

n nG

n nkl kl x nkl ln

l k ll

nG

in klinl k

� �

� �

�

�

�

�

�

�

�

� ��

� ��

� ��

�2 2

2exp , 1, 2, ...

in n tx n

l kl

� ��

Using the fact that cosh ( ) cos ,in n� �� and

sinh sinin n

x i xl l

� ��

the above expression becomes

2 2 2 2

2 2 2

2 ( 1)( , ) sin exp , 1, 2, ...

nn n n nu x t x G t n

ll k kl kl

� � � ��


2 22 2 2

2 21

2( , ) ( 1) sin exp[( / ) ]n

n

n nu x t nG x n kl t

ll k kl

π π π π=

⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠∑�

(6.61)

EXAMPLE 6.34 Find the solution of the BVP given by

2

2

1 ,PDE:u u

k tx�

� ��

0 , 0x a t� � �

BCs: (0, ) ( ),u t f t= ( , ) 0u a t �

IC: ( , 0) 0u x =

using the Laplace transform method.



Solution Taking the Laplace transform of the given PDE, we have

2

2

1[ ( , ) ( , 0)]

d UsU x s u x

kdx= − (6.62)

Using the initial condition ( , 0) 0,u x � we get

2

2( , ) 0

d U sU x s

kdx− = (6.63)


( / ) ( / )( , ) s k x s k xU x s Ae Be−= + (6.64)

Now taking the Laplace transform of the BCs, we have

(0, ) ( )U s F s� (6.65)

( , ) 0U a s � (6.66)

Using Eqs. (6.64) and (6.65), we obtain

( )F s A B� � (6.67)


( , ) exp [ ( / ) ] exp [ ( / ) ]U a s A s k a B s k a= + − (6.68)

Combining Eqs. (6.67) and (6.68), we get

2 /( ) (1 )a s kF s A e= −Therefore,

2 /( )/(1 )a s kA F s e= −and

2 /

2 / 2 /

( ) ( )( )

(1 ) 1

a s k

a s k a s k

F s F s eB F s

e e= − −

− −

or

/ / /( ) /( )a s k a s k a s kA F s e e e− −= −

and

/ / /( ) /( )a s k a s k a s kB F s e e e−= −

Therefore,

{ }/ ( ) / ( )/ /

( )( , )

( )

s k a x s k a x

a s k a s k

F sU x s e e

e e

− − −−

= −−



sinh ( )( )

sinh

sa x

kF s

sa

k

� ��

� � �

(6.69)

Its inverse Laplace transform yields

1( ) sinh ( )

( , ) ;

sinh

sF s a x

ku x t L t

sa

k

�

� ��

� � � ��

(6.70)

( ) sinh ( )1

2sinh

sti

i

se F s a x

kds

i sa

k

�

��

�

�

� ��

� � �

��

�(6.71)

��

EXAMPLE 6.35 Using the Laplace transform method, solve the IBVP described as

2

1PDE: cos ,xx ttu u t

cω= − 0 ,x� � � 0 t� � �

BCs: (0, ) 0,u t = u is bounded as x tends to �

ICs: ( , 0) ( , 0) 0tu x u x= =

Solution Taking the Laplace transform of PDE, we obtain

22

2 2 2 2

1[ ( , ) ( , 0) ( , 0)]t

d U ss U x s su x u x

dx c s �

� � � �

�

Using the ICs, we get

2 2

2 2 2 2( , )

d U s sU x s

dx c s �

� � �

�

(6.72)


2( / ) ( / )

2 2( , )

( )s c x s c x c

U x s Ae Bes s ω

−= + ++



As ,x�� the transform should also be bounded which is possible if 0;A � thus,

2( / )

2 2( , )

( )s c x c

U x s Bes s ω

−= ++

(6.73)

Taking the Laplace transform of the BC, we get

(0, ) 0U s �

Using this result in Eq. (6.73), we have

2

2 2( )

cB

s s �

� �

�

Hence,

2( / )

2 2( , ) [1 ]

( )s c xc

U x s es s ω

−= −+

(6.74)

Now, taking its inverse Laplace transform, we get

( / )2 1 2 1

2 2 2 2

1( , ) ; ;

( ) ( )

s c xeu x t c L t c L t

s s s sω ω

−− − ⎡ ⎤⎡ ⎤= − ⎢ ⎥⎢ ⎥+ +⎢ ⎥⎣ ⎦ ⎣ ⎦

(6.75)

But,

1 1 12 2 2 2 2 2

1 1 1 1; ; ; (1 cos ),

( )

sL t L t L t t

ss s s�

� � � �

� � �

� ��

( / )1

2 2 2

1; 1 cos ,

( )

x c se x xL t t H t

c cs sω

ω ω

−− ⎡ ⎤ ⎧ ⎫⎛ ⎞ ⎛ ⎞= − − −⎨ ⎬⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭+⎢ ⎥⎣ ⎦

where H is the Heaviside unit function. Substituting these results in Eq. (6.75), we arrive at

2 2

2 2( , ) (1 cos ) 1 cos

c c x xu x t t t H t

c cω ω

ω ω⎡ ⎤⎧ ⎫⎛ ⎞ ⎛ ⎞= − − − − −⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎩ ⎭⎣ ⎦

EXAMPLE 6.36 Solve the IBVP described by

PDE: , 0 1, 0tt xxu u x t= < < >

BCs: (0, ) (1, ) 0, 0u t u t t= = >

ICs: ( , 0) sin , ( , 0) sin , 0 1tu x x u x x xπ π= = − < <

Solution Taking the Laplace transform of the given PDE, we get

22

2( , ) ( , 0) ( , 0)t

d Us U x s su x u x

dx� � �



Using the initial conditions, this equation becomes

22

2(1 ) sin

d Us U s x

dxπ− = − (6.76)


2 2

( 1) sin( , ) sx sx s x

U x s Ae Bes

�

�

�

��

�

(6.77)

The Laplace transform of the BCs gives

(0, ) 0, (1, ) 0U s U s� � (6.78)

Using Eq. (6.78) into Eq. (6.77), we find 0.A B� � Hence, we obtain

2 2

( 1) sin( , )

s xU x s

s

�

�

��

�

(6.79)

Taking the inverse Laplace transform, we get

1 1 12 2 2 2 2 2

1 1( , ) sin ; sin ; ;

sinsin cos

s su x t x L t x L t L t

s s s

tx t

� �

� � �

�

� �

�

� � �

� � � � ��

� ��

Hence the required solution of the given IBVP is

sin( , ) sin cos

tu x t x t

��

�

� ��

EXAMPLE 6.37 If the function ( , )u x t satisfies the following:

2

1PDE: , 0 , 0xx ttx u k x l t

c= + ≤ ≤ >

BCs: (0, ) ( , ) 0,xu t u l t= = 0,t �

ICs: ( , 0) ( , 0),tu x u x= 0 x l� �

then, show that

21

3

cosh { ( )/ }( , ) 1 ;

cosh { / }

kc s l x cu x t L t

sl cs− ⎡ ⎤−⎛ ⎞= −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

and find the solution.



Solution Taking the Laplace transform of the given PDE, we get

22

2 2

1[ ( , ) ( , 0) ( , 0)]t

d U ks U x s su x u x

sdx c� � � �

Using the initial conditions, we obtain

2 2

2 2( , )

d U s kU x s

sdx c− = (6.80)


2( / ) ( / )

3( , ) s c x s c x kc

U x s Ae Bes

−= + − (6.81)

Taking the Laplace transform of the BCs, we get

(0, ) ( , ) 0dU

U s l sdx

= = (6.82)

Using these boundary conditions, Eq. (6.81) gives

2 3/A B kc s+ =

( / ) ( / ) 0s c l s c lAs se B e

c c−− =

Eliminating B, we get

( / ) ( / ) ( / )2

[ ]s c l s c l s c ls kcA e e e

c s− −+ =

which gives

2( / )

3/cosh

2s c lkc s

A e lcs

− ⎛ ⎞= ⎜ ⎟⎝ ⎠

2( / )

3/cosh

2s c lkc s

B e lcs

⎛ ⎞= ⎜ ⎟⎝ ⎠

Hence, from Eq. (6.81), we have

2 2

3 3

cosh ( )( , )

cosh

sl x

kc kccU x sss slc

� ��

(6.83)



Applying the complex inversion formula, we obtain

2 1 2 13

3

cosh ( )1

( , ) ; ;cosh

sl x

cu x t kc L t kc L ts ss lc

− −

⎡ ⎤⎡ ⎤−⎢ ⎥⎢ ⎥ ⎡ ⎤⎣ ⎦= −⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎣ ⎦⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(6.84)

2 2 2

3

cosh ( )

2 2cosh

sti

i

se l x

kc kc tc dssi

s lc

�

��

�

�

� ��

��

�(6.85)

But

3

(2 / ) (2 )

cosh ( )1

sum of the residues at the poles 0 of order2 three and at the poles of cosh ( / ) , i.e., atcosh

1

sti

i

s c l i n

se l x

c ds ssi s c ls lc

e e

γ

γ

π π

π+

−

+

⎡ ⎤−⎢ ⎥⎣ ⎦ = =⎛ ⎞⎜ ⎟⎝ ⎠

= − =

∫�

�

or at

(2 1), 0, 1, 2, ...

2

n c is n

l

��

Now the residue at the pole s = 0 of order 3 is

2

20 0

2

cosh ( ) cosh ( )1 1

Lt Lt2 2

cosh cosh

sinh ( ) cosh ( ) sinh

cosh cosh

st

st xt

s s

st

s se l x l x

d d l xc cte e

s sds cds l lc c

s s sl x l x l

lc c ce

s scl l

c c

� �

� � ��

� � � � � ��

� ��

�

� � � �

Again differentiating and taking the limit, we get

2 2 2 22

2 2 2 2

1 ( ) 2

2 2 2 2

l x l t x lxt

c c c c

� ��

�



Also, the residue at the poles

{ }

3

3 3 3 3

3 3

,(2 1)

, 0,1, 22

1 1exp ( ) cosh , 0,1, 2, ...

sinh

2 1 2 1exp cosh

2 2

(2 1) (2 1)sinh

22

n

n nn

n

i n cs n

l

l xs t s n

l l cs sc c

n n l xc ict ic

l l c

n i c n ll ic

l cl

π

π π

π π

…+= =

−⎛ ⎞= =⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠

⎧ + ⎫ ⎧ + − ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎨ ⎬⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭⎩ ⎭=

+ +

�

� �

2

3 3 3

2

3 3 2

2 1 2 18 exp cosh ( )

2 2

(2 1)(2 1) ( ) sinh

2

2 1 2 18 exp cos ( )

2 2(since sinh sincosh cos )(2 1) sin (2 1)

2

n nl c c ict i l x

l l

nn i c i

n nl i ct l x

l li i

in c n

� �

� �

� �

� ��

� ��

� � � � � ��

� �

� � � � � ��

Its real part is

2

3 3 2

2 1 2 18 cos cos ( )

2 2, 0,1, 2, ...

( 1) (2 1)n

n nl ct l x

l ln

n c

π π

π

+ ⎧ + ⎫⎛ ⎞ ⎛ ⎞ −⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎩ ⎭ =− +

Therefore,

2 21

2 23

2

3 3 20

cosh ( )

2 2cosh

2 1 2 1( 1) 8 cos cos ( )

2 2

(2 1)

n

n

sl x

t x lxcL

s c cs lc

n nl ct l x

l l

n c

� �

�

�

�

� ��

� � � ��

��



Thus, the required solution from Eq. (6.84) is

2

3 30

8 ( 1) 2 1 2 1( , ) ( 2 ) cos cos ( )

2 2 2(2 1)

n

n

kx kl n nu x t x l ct l x

l ln� �

��

� � � � � � ��

EXAMPLE 6.38 A string is stretched and fixed between two points (0, 0) and (l, 0). Motion

is initiated by displacing the string in the form sin ( / )u x l� �� and released from rest at time

0.t � Find the displacement of any point on the string at any time t.

Solution The displacement ( , )u x t of the string is governed by

2PDE: ,tt xxu c u= 0 ,x l< < 0t �

BCs: (0, ) ( , ) 0u t u l t= =

ICs: ( , 0) sin ( / ),u x x lλ π= ( , 0) 0tu x �

Taking the Laplace transform of the given PDE, we have

22 2

2( , ) ( , 0) ( , 0)t

d Us U x s su x u x c

dx� � �

Using the ICs, we get

2 2

2 2 2sin

d U s U s x

ldx c c

� �� (6.86)


( / ) ( / )2 2 2 2

sin ( / )( , )

/s c x s c x s l x

U x s Ae Bes c l

λ ππ

−= + ++

(6.87)

The Laplace transform of the BCs is given by

(0, ) 0, ( , ) 0U s U l s� �

Applying these conditions in Eq. (6.87), we obtain

/ /

0

0sl c sl c

A B

Ae Be−

+ =

+ =

On solving the above set of equations, we get only the trivial solution, viz.

0A B� �

Thus,

2 2 2 2

sin ( / )( , )

/

s l xU x s

s c l

λ ππ

=+



Its inverse Laplace transform yields

12 2 2 2

( , ) ; sin cos sin/

s x c xu x t L t t

l l ls c l

π π πλ λπ

− ⎡ ⎤ ⎛ ⎞= = ⎜ ⎟⎢ ⎥ ⎝ ⎠+⎣ ⎦

which gives the displacement of the string for a given time.

EXAMPLE 6.39 An infinitely long string having one end at 0x � is initially at rest on the

x-axis. The end 0x � undergoes a periodic transverse displacement described by 0 sin , 0.A t tω >Find the displacement of any point on the string at any time t.

Solution Let ( , )u x t be the transverse displacement of the string at any point x and at

any time t. Then the transverse displacement of the string is described by the

2PDE: , 0, 0tt xxu c u x t= > >

0 0BCs: (0, ) sin , 0 ( constant)u t A t t Aω= > =

ICs: ( , 0) 0, ( , 0) 0, 0tu x u x x= = ≥The Laplace transform of the PDE gives

22 2

2( , 0) ( , 0)t

d Us U su x u x c

dx� � � (6.88)

After using the ICs, Eq. (6.88) becomes

2 2

2 2( , ) 0

d U sU x s

dx c� � (6.89)


( / ) ( / )( , ) s c x s c xU x s Ae Be−= +

Since the displacement ( , )U x s is bounded as ,x�� we get A = 0 and, therefore,

( / )( , ) s c xU x s Be−= (6.90)

Now taking the Laplace transform of the BC, we obtain

02 2

(0, )A

U ss

�

�

�

�

Using this expression, from Eq. (6.90), we have

02 2

AB

s

�

�

�

�



Hence the solution of Eq. (6.89) is found to be

( / )02 2

( , ) s c xAU x s e

s

ωω

−=+

(6.91)

Finally, taking the inverse Laplace transform of Eq. (6.91), we get

1 ( / )02 2

0

( , ) ;

sin if

0 if

s c xAu x t L e t

s

x xA t t

c cx

tc

ωω

ω

− −⎡ ⎤= ⎢ ⎥+⎣ ⎦

⎧ ⎛ ⎞− >⎜ ⎟⎪⎪ ⎝ ⎠= ⎨⎪ <⎪⎩

�� %��

EXAMPLE 6.40 Find the Laplace transform of erf 1

.t

� ��

Solution Following the definition of Laplace transform, we have

2

2

1/

0 0

1/

0 0

2[erf (1/ ); ]

2

tst u

t st u

L t s e e du dt

e du dt

�

�

� �

� �

�

�

� �

� �

�

�

Now, changing the order of integration (see Fig. 6.5), we get

22

2 2 2

1/

0 0

/

0

2[erf (1/ ); ]

2( )

uu st

u s u u

L t s e du e dt

e e dus

�

�

� �

� � �

�

� � �

� �

�

�

�(6.92)

To evaluate this integral, consider an integral of the form

2 2 2/

0( , ) a u b uI a b e du� �

� ��

(6.93)

We can easily verify that

2 2

0( , 0)

2a uI a e du

a

��

� ��

(6.94)



t

u

O

u t = 1/��

Fig. 6.5 Curve of 1/ .t

since

2

0

21ue du

�

�

��

Differentiating Eq. (6.93) partially with respect to b, we get

2 2 2 22 /

02 a u b uIb u e du

b� � �

� � ��

�(6.95)

Let ; then / ,au x du dx a= = and we can easily verify that

2 2 2 2

0

1 1( , ) exp ( / ) (1, )I a b x a b x dx I ab

a a� � � ��

�(6.96)

Also, let / ;b u x= then we find from Eq. (6.95) that

2 2 20 22 2 2

2 2 2

20

( , )2 exp

2 exp

2 (1, )

I a b x a b bb x dx

b b x x

x a bdx

x

I ab

�

�

� ��

� ��

� �

�

�

��

(6.97)

From Eqs. (6.96) and (6.97), we can eliminate (1, )I ab and arrive at

2I

aIb� ��

�



On integration, we get2ln 2 ln , ( , ) abI ab c I a b ce��

Thus,

2( , ) ( , 0) abI a b I a e��

2 2 2 2/ 2

0 2a u b u abe du e

a

��

��

(6.98)

By making use of Eq. (6.98), the Laplace transform of the given function is obtained fromEq. (6.92) as

2 22 1[erf (1/ ); ] (1 )

2 2s sL t s e e

s s

π ππ

− −⎛ ⎞= − − = −⎜ ⎟⎝ ⎠

EXAMPLE 6.41 Find the inverse Laplace transform of 1/

.se

s

−

Solution The given function can be rewritten as

1/1 2

0 0

1 1 ( 1) ( 1)

! !

n ns

n nn n

es s n s n s

�

�

� �

� �

� ��

Therefore,

1/ 1/21 1

1/20 0

( 1) 1 ( 1); ;

! ! 1/2

s n n n

nn n

e tL t L t

s n n ns

� �

� �

�

� �

� � � ��

But,

2

1 (2 )!

2 2 !n

nn

nπ+ =

Hence,1/ 2 1/2

1

0

1/2 1/2 4 3/2

2 4

( 1) 2;

(2 )!

4 2

2! 4!

1 (2 ) (2 )1

2! 4!

cos 2

s n n n

n

e tL t

s n

t t t

t t

t

t

t

�

� � �

�

�

� �

�

�

�

� � ��

�

� �

� ��

�

�

�

�

�



EXAMPLE 6.42 Solve

sin , cosx ky a kt y kx a kt� � � ��

using Laplace transform technique, given that

(0) 0, (0)x y b� �

Solution Taking the Laplace transform of the given equations, we get

2 2(0)

aksX x kY

s k− + =

+(6.99)

2 2(0)

assY y kX

s k− − =

+(6.100)

Utilizing the given ICs, we obtain

2 2

aksX kY

s k+ =

+(6.101)

2 2

assY kX b

s k� � �

�

(6.102)

Solving Eqs. (6.101) and (6.102) and using Cramer’s rule, we obtain

2 2

2 2

2 2

akk

s k bks kXas k s s kb s

s k

��

� ��

�


2 2 2 2

a bsY

s k s k= +

+ +Taking the inverse Laplace transform of the above two equations, we get

sin , sin cosa

x b kt y kt b ktk

� � � �

which is the required solution.

EXAMPLE 6.43 A beam which is coincident with the x-axis is simply supported at its end

0x � and is clamped at the other end .x l� Let a vertical load W act transversely on the

beam at /4.x l= The differential equation for deflection at any point is given by

4

4( /4)

d y Wx l

EIdxδ= −

where EI is the flexural rigidity of the beam. Find the deflection at any point.



Solution Taking the Laplace transform on both sides of the governing differentialequation and noting the transform of Dirac delta function from Eq. (6.23), we obtain

4 3 2(0) (0) (0) (0) exp [ ( /4) ]W

s Y s y s y sy y l sEI

′ ′′ ′′′− − − − = − (6.103)

Since the beam is simply supported at 0,x � we have

(0) (0) 0y y�� (6.104)

Also, it is given that the beam is clamped at ,x l� which means that

( ) ( ) 0y l y l�� (6.105)

Let (0) and (0) .y A y B� �� Then using Eq. (6.104) into Eq. (6.103), we get

4 2 ( /4)l sWs Y As B e

EI−− − =

( /4)

2 4 4

l sA B W eY

EIs s s

−= + +


31 ( /4)

4

1{ } ;

3!l sBx W

y Ax L e xEI s

− −⎡ ⎤= + + ⋅⎢ ⎥⎣ ⎦(6.106)

But, from the second shifting property (Theorem 6.14), we have

1 ( /4)4

1{ } ; ( /4) ( /4)l sL e x f x l H x l

s− −⎡ ⎤⋅ = − −⎢ ⎥⎣ ⎦

where

31

4

1( ) ;

3!

xf x L x

s�

� ��

� �

Hence Eq. (6.106) becomes

33 ( /4)

( /4)3! 3!

B W x ly Ax x H x l

EI

��

Thus, the deflection is given by

3

33

for 0 /46

( /4)for /4

6 6

By Ax x x l

B W x lAx x l x l

EI

= + < ≤

−= + + < <



The unknowns A and B can now be determined by using the conditions (6.105). Thus, wehave

33 9

06 128

B WlAl l

EI= + +

2

2 90

2 32

B WlA l

EI= + +

whose solution gives

29 81,

256 128

Wl WA B

EI EI= = −

Hence the resulting deflection y is given as

3 3

33

9 27 , 0 /4256 256

18( /4) , /4

256 6

Wl Wlx l

EI EIy

Wl Wx l l x l

EI EI

⎧− < ≤⎪

⎪= ⎨⎪− + − < <⎪

⎩

EXAMPLE 6.44 Find the solution of the

PDE: , 0 , 0t xxu ku x t� � � ��

subject to

BCs: ( , ) 0 as

(0, ) ( )

IC: ( , 0) 0

u x t x

u t g t

u x

� �

�

�

�

assuming

21

3exp ; exp

42

s x xL x t

k ktk t�

�

� ��

Solution Taking the Laplace transform of the PDE, we have

2

2( , ) ( , 0)

d UsU x s u x k

dx� � (6.107)



Using IC: ( , 0) 0,u x = we obtain

2

2( , ) 0

d U sU x s

kdx� � (6.108)

whose general solution is given by

( , ) exp exps s

U x s A x B xk k

� � � ��

� �

The Laplace transform of the first BC gives:

0 asU x� � �

Using this in the solution, we get 0A � and

( , ) exps

U x s B xk

� ��

� �(6.109)

The Laplace transform of the second BC gives

(0, ) ( )U s G s=

Using this condition, Eq. (6.109) becomes

( , ) ( ) exps

U x s G s xk

� ��

� �(6.110)

Taking inverse Laplace transform, we get

1 1 23

( , ) ( ) exp ; [ ( ); ] exp( /4 );2

s xu x t L G s x s L L g t s L x kt s

k k t��

� � � ��

Finally, through the use of the convolution theorem, we arrive at the result

2

3/20

exp[ /4 ( )]( , ) ( )

2 ( )

t x x k t uu x t g u du

k t uπ− −=

−∫

EXAMPLE 6.45 Find the solution of IBVP described by

PDE: ( , ) ( , ) ( , ),

constant, 0 , 0

BCs: (0, ) 0, 0

( , ) 1, 0

IC: ( , 0) 0, 0

t xxu x t u x t hu x t

h x t

u t t

u t t

u x t

π

π

= −

= < < >

= >

= >

= =



Solution Taking the Laplace transform of the PDE with respect to the variable t, wehave

2

2( , ) ( , 0) ( , )

d UsU x s u x hU x s

dx� � � (6.111)

Using the IC: ( , 0) 0,u x = we get

2

2( ) ( , ) 0

d Uh s U x s

dx� � � (6.112)


( , ) cosh ( ) sinh ( )U x s A h s x B h s x= + + + (6.113)

Taking the Laplace transform of BCs, we obtain

(0, ) 0, ( , ) 1/U s U s s�� (6.114)

Using these BCs into Eq. (6.113), we have

( , ) sinh ( )

1sinh ( )

U x s B h s x

B h ss

π

= +

= +

implying thereby

1 1

sinh ( )B

s h sπ=

+

Hence, the required solution of Eq. (6.112) is

1 sinh ( )( , )

sinh ( )

h s xU x s

s h sπ+=+

By means of complex inversion formula, we get

1 sinh ( )( , )

2 sinh ( )

sum of the contributions from all the poles of the integrand

sti

i

e h s xu x t

i s h s

�

��

�

�

��

�

�

��

�

The poles are given by 0, and sinh ( ) 0 sin ( ).s h s i h sπ π= + = = + Therefore,

2, implyingi h s n h s nπ π+ = + = −



Thus, the integrand has poles at 0,s � and

2 , 1, 2, 3,s n h n= − − = …

The residue of the expression

sinh ( )at 0

sinh ( )

ste h s xs

s h sπ+ =+

is

sinh ( )

sinh ( )

h x

hπ

The residue of the integrand at 2s n h� � � is

2( ) 2

22 2

2

sinh ( ), 1, 2, 3, ...

( )sinh cosh ( )

2

n h te n xn

n hn n

n

ππ π

− − − =− −− + −

−

Using the relations sinh sin , and cosh cos ,x i ix x ix� � the above residues become

2 2

2 2

2

2

exp [ ( ) ] sin 2 exp [ ( ) ]sin

( ) ( ) coscos

2

2 exp [ ( ) ]sin

( ) ( 1)n

n h t i nx n n h t nx

n h n h nn

in

n n h t nx

n h

� � ��

�

� � � ��

� � �

� ��

� �


2

21

sinh ( ) 2 ( 1) exp ( )sin( , )

sinh ( )

ht n

n

h x e n n t nxu x t

h n hπ π

−

=

− −= ++∑

�

�%��

1. Find the Laplace transform of the following:

(i) cos ,t at (ii) sin ,tte t�

(iii) sin 2 ,tte t (iv) 2(1 ) att e−+ .



2. Find the Laplace transform of ( )f t defined as

(i)/ , 0

( )1,

t tf t

t

τ ττ

< <⎧= ⎨ >⎩

(ii)

, 0 1

( ) (1 ), 1 2

0, 2

t t

f t t t

t

< <⎧⎪= − < <⎨⎪ >⎩

3. Find the Laplace transform of

(i) 1,

te

t

− (ii) cos 2 sin 3t t

t

� .

4. Obtain the Laplace transform of

, 0 1( )

0, 1 2

t tf t

t

� ��

� ��

given ( 2) ( ) for 0.f t f t t� � �


, 0 6( )

12 , 6 12

t tf t

t t

� ��

� � ��

given ( 12) ( ) for 0.f t f t t� � �


1, 0( )

1, 2

t bf t

b t b

� ��

� � ��

given ( 2 ) ( ).f t b f t� �

7. Find the Laplace transform of erfc (1/ ).t


(i) 1( ),J t (ii) 1( ).tJ t

9. Find the inverse Laplace transform of

(i) 1 ,( 1)( 2)s s s+ +

(ii)3( 2)

s

s �

(iii)2

1,

( 1)s s �

(iv)2

4 5 ,( 1) ( 2)

s

s s

+− +

(v)2 2

3 ,( 6 13)

s

s s

++ +

(vi)2

3 2

1 .3 2

s

s s s

++ +




(i) ,lns a

s b

++

(ii) 1 ,tans

k− ⎛ ⎞⎜ ⎟⎝ ⎠

(iii)2

2

1 ,ln( 1)

s

s

+−

(iv)2

4 .ln4

s

s

−+

11. State and prove the convolution theorem for Laplace transforms.12. Using the convolution theorem, find the inverse Laplace transform of

(i)2 2 2

1 ,( )s s a+

(ii) 1 ,( )s s a−

(iii)2 3

.( 4)

s

s +

13. Find the inverse Laplace transform of2/( 1)se sπ− +

14. Using the Heaviside expansion theorem, find the inverse Laplace transform of

(i)3

1 ,1s +

(ii)2

3 1 .( 1)( 1)

s

s s

+− +


cosh ( ) ,( ) 0 1cosh

x sF s x

s s= < <

using the complex inversion formula.16. Solve the following ODE using the Laplace transform

23 2 4 ty y y e��

given that (0) 3, (0) 5.y y′= − =17. Solve the ODE by the Laplace transform method

44

40

d yk y

dt� �

with the initial conditions (0) 1, (0) (0) 0, (0) 0.y y y y� ��

18. Solve the initial value problem using the method of Laplace transform

0, (0) 1, (0) 0y ty y y y��

19. Using the Laplace transform method, solve the system of equations

1

2

dx dyx y

dt dt

dyx y

dt

� � � �

� �

given that 0 and 1 when 0.x y t� � �



20. Applying the Laplace transform technique, solve the system:

0

2 2 2 0

tdx dyx e

dt dt

dx dyx y

dt dt

�

� � � �

� � � �

given that 1 and 1 when 0.x y t� � � �

21. Using the Laplace transform, find the solution of the system of ODEs

2 2

2 21, 0

d x d xy x

dt dt� � � �

satisfying the ICs (0) (0) (0) (0) 0x y x y′ ′= = = = .

22. Show that, by means of the Laplace transform, the solution ( , )x t� of one-dimensionaldiffusion equation

2

2

1, 0 , 0x t

k tx

θ θ π� � � ��

satisfying the boundary conditions

(0, ) 1 , 0

( , ) 0, 0

( , 0) 0, 0, 0

tt e t

t t

x t x

θ

θ π

θ π

−= − >

= >

= = ≤ ≤

is given by the formula

2

21

sin {( ) / } 2 sin exp ( )( , )

sin( / ) ( 1)

t

n

x e x k nx n ktx t

k n n k

� ��

� � �

�

�

� � �

� � �

��

23. Using the Laplace transform method, solve

PDE: 3

BCs: , 0, (0, ) 02

IC: ( , 0) 30 cos 5

t xx

x

u u

u t u t

u x x

π

=

⎛ ⎞ = =⎜ ⎟⎝ ⎠

=

24. Using the Laplace transform, solve the following problem in wave propagation:

2PDE: , 0 , 0

BCs: 0 at 0

at , 0 ( , are constants)

ICs: ( , 0) 0, 0

xx tt

t

t

c u u x l t

u x

Eu P x l t E P

u u x x l

= < < >

= =

= = >

= = < <



25. Solve the BVP using the Laplace transform method

2PDE: , 0 , 0

BC: (0, ) 0

ICs: ( , 0) 0,

( , 0) 1

tt xx

t

u c u x t

u t

u x

u x

� � � � �

�

�

�

� �

26. If � is the potential, i the current, l the inductance per unit length of cable, c the

capacitance to ground per unit length, then � satisfies the wave equation

xx ttlc� ��

Initially, the line is considered to be dead, i.e.

( , 0) ( , 0) 0tx xφ φ= =

The other boundary conditions are

0(0, ) ( ),

( , ) ( , ) 0, 0x

tt f t

l t l t t

φ

φ φ

>=

= = >

Find the potential at any point on the cable at any time t, assuming .l � �


��

Joseph Fourier, a French mathematician, had invented a method called Fourier transform in1801, to explain the flow of heat around an anchor ring. Since then, it has become a powerfultool in diverse fields of science and engineering. It can provide a means of solving unwieldyequations that describe dynamic responses to electricity, heat or light. In some cases, it canalso identify the regular contributions to a fluctuating signal, thereby helping to make senseof observations in astronomy, medicine and chemistry. Fourier transform has become indispensablein the numerical calculations needed to design electrical circuits, to analyze mechanicalvibrations, and to study wave propagation.

Fourier transform techniques have been widely used to solve problems involving semi-infinite or totally infinite range of the variables or unbounded regions. In order to deal withsuch problems, it is necessary to generalize Fourier series to include infinite intervals and tointroduce the concept of Fourier integral. In this chapter, we deal with Fourier integralrepresentations and Fourier transforms together with some applications to Diffusion, Waveand Laplace equations.

��

Definition 7.1 (Dirichlet’s conditions). A function ( )f x is said to have satisfied Dirichlet’s

conditions in the interval ( , ),L L� provided ( )f x is periodic, piecewise continuous, and has

a finite number of relative maxima and minima in ( , ).L L�

Let a function ( )f x be periodic with period 2 , i.e., ( 2 ) ( ),L f x L f x� � and satisfy

Dirichlet’s conditions in the interval ( , ).L L� Then f (x) has a Fourier series representation for

every x in the form

388

��

Fourier Transform Methods


FOURIER TRANSFORM METHODS 389

0

1

( ) cos sin2 n n

n

a n x n xf x a b

L L

π π�

�

� �� (7.1)

where

1( )cos , 0,1, 2, ...

�

�

� ��L

nL

n ta f t dt n

L L(7.2)

1( )sin , 1, 2, ...

�

�

� ��L

nL

n tb f t dt n

L L(7.3)

Here, ,n na b are called Fourier coefficients. Fourier series representation, however, can be

extended to some non-periodic functions also, provided the integral of the modulus of such

a function ( )f t satisfies the condition

| ( ) |f t dt�

��is finite.

Substituting Eqs. (7.2) and (7.3) into Fourier series (7.1), we get

� � �1

1 1 1( ) ( ) ( ) cos cos ( )sin sin

2

L L L

L L Ln

n t n x n t n xf x f t dt f t dt f t dt

L L L L L L L

π π π π�

� � ��

�� Noting that cos ( ) cos cos sin sin ,A B A B A B� � � and interchanging the order of summation

and integration, we obtain

1

1 1 ( )( ) ( ) ( ) cos

2

L L

L Ln

n t xf x f t dt f t dt

L L L

π�

� � �

�� (7.4)

Further, if we assume that the function ( )f x is absolutely integrable, and allowing L to tend

to infinity, i.e.,

| ( ) |f t dt�

�� (7.5)

we get

1Lt ( ) 0

2

L

LLf t dt

L �� (7.6)

In the remaining part of the infinite sum of Eq. (7.4), if we set / ,s LπΔ = the equation reduces to

01

1( ) Lt ( ) cos { ( )}

s

ssn

f x f t n s t x dt�

��

��

� ��

� � �� (7.7)



As , 0,L sΔ�� implying that sΔ is a small positive number and the points n sΔ are

equally spaced along the s-axis. The series under the integral can be approximated by an

integral of the form (as 0)sΔ �

0cos { ( )}s t x ds

��

Thus, Eq. (7.7) can be rewritten as

0

1( ) ( ) cos { ( )}f x f t s t x ds dt

π� �

�� (7.8)

0

1( ) cos { ( )}f t s t x dt ds

π� �

�� (7.9)

which is the Fourier integral representation of ( ).f x

�� !

Theorem 7.1 If ( )f x satisfies Dirichlet’s conditions for x�� and if the integral

( )f x dx�� is absolutely convergent, then

0

1 1( )cos ( ) [ ( 0) ( 0)]

2d f t t x dt f x f xα α

π� �

�� (7.10)

To establish this result, the central results required are the Riemann-Lebesgue lemma and theRiemann localization lemma; first we shall state and prove the former.

Riemann-Lebesgue lemma If ( )f x satisfies Dirichlet’s conditions in the interval (a, b),

then each of the integrals

( )sin , ( )cos� �b b

a af x Nx dx f x Nx dx

tends to zero as .N � �

Proof Suppose 1 2, , ..., pa a a are the points in ( , )a b taken in the order at which the

function ( )f x has either a turning value or a finite discontinuity. Let 0 1, and .pa a b a�

� �

Then we may write

1

0

( )sin ( )sinr

r

pb a

a ar

f x Nx dx f x Nx dx�

�

�� (7.11)



Now in each of the sub-intervals 1( , ), 0,1, 2, ..., , ( )r ra a r p f x�

� is a continuous function and

is either monotonically increasing or decreasing. Thus, aplying the second Mean Value Theoremof integral calculus to each of these intervals, we have

1 11( )sin ( ) sin ( ) sin

r r

r r

a a

r ra a

f x Nx dx f a Nx dx f a Nx dx�

�

� �

�� (7.12)

where � is some value of x in the range 1( , ).r ra a + Carrying out the integrations on the right-

hand side of Eq. (7.12), we get

11

cos cos cos cos( ) ( )r r

r rNa N N Na

f a f aN N

� ��

�

� ��

Now taking the limit as ,N �� we get

1Lt ( ) sin 0

r

r

a

aNf x Nx dx

�

��

Since the number of terms on the right-hand side of Eq. (7.11) is finite, interchanging ofsummation and limit process can be carried out and, therefore,

1 1

0 0

Lt ( ) sin Lt ( )sin Lt ( )sin 0r r

r r

p pb a a

a a aN N Nr r

f x Nx dx f x Nx dx f x Nx dx� �

��

i.e.,

Lt ( ) sin 0b

aNf x Nx dx

�� (7.13)

which is the Riemann-Lebesgue lemma.

Riemann localization lemma If ( )f t satisfies Dirichlet’s conditions in the interval (0, a),

where a is finite, then

0

sin( ) (0 )

2

a Ntf t dt f

t

π→ +∫as .N � �

Proof We may write

1 20 0 0

sin sin sin( ) (0 ) { ( ) (0 )}� � � � � � ��

a a aNt Nt Ntf t dt f dt f t f dt I I

t t t



Since the function ( )f t� is continuous in (0, a), from the definition of derivative

( ) (0 )f t f

t

� �

is continuous in (0, a). By the Riemann-Lebesgue lemma (since the integrand of 2I is

bounded as 2) 0 as .N I N� � � � � Hence,

0 0 0

0

sin sin sinLt ( ) (0 ) (0 )

sin(0 ) (0 )

2

a a Na

N

Nt Nt uf t dt f dt f du

t t u

uf du f

u

π

��

�

� � � �

� � � �

� � �

� (7.14)

which is the Riemann localization lemma.

Proof (Fourier integral theorem). Since the integral ( )f x dx�� is absolutely convergent,

| ( )|f x dx�� is finite and converges for all � in the range (0, N). Also, | cos ( ) | 1,t x� � �

implying that the integral

( ) cos ( )f t t x dtα�

��

converges and is independent of and .x� Thus, after changing the order of integration, the

double integral.

0( ) cos ( )

NI f t t x dt dα α

�

��

� �� (7.15)

can be expressed as

0

sin ( )( ) cos ( ) ( )

N N t xI f t t x d dt f t dt

t xα α

� �

��

�� Let .v t x� � Then the above integral becomes

0

1 2 3 40

sin sin( ) ( )

Nv NvI f v x dv f v x dv I I I I

v v

δ δ

δ δ

� � �

��

� � � � � � � � � � � �� When 1 4, andN I I� � both tend to zero in view of the Riemann-Lebesgue lemma. Thus the

only contribution to the integral will be from the neighbourhood of 0.v � Using the Riemannlocalization lemma, we get



30

sinLt ( ) ( )

2

� �

��

� � � ��N

NvI f v x dv f x

v(7.16)

and the second integral

0

20

sin sin( ) ( ) ( )

2

Nv NvI f v x dv f x v dv f x

v v

δ

δ

π−

= + = − = −∫ ∫ (7.17)

Incorporating these results into Eq. (7.15), we obtain

0( ) cos ( ) [ ( ) ( )]

2f t t x dt d f x f x

πα α� �

��

� �� (7.18)

If f is a continuous function of x, then

( ) ( ) ( )f x f x f x� � � �

and Eq. (7.18) reduces to

0

1( ) ( ) cos ( )f x d f t t x dtα α

π� �

��

� �� (7.19)

If x is a point of discontinuity, then

1( ) [ ( ) ( )]

2f x f x f x� � � � (7.20)

i.e., the intergral (7.19) converges to the average value of the right- and left-hand limits. Thus,the proof of the Fourier integral theorem is complete.

In order to bring out the analogy between Fourier series and Fourier integral theorem, werewirte Eq. (7.19) as

0

1( ) ( ) (cos cos sin sin )f x f t t x t x dt dα α α α α

π� �

��

If we define

1( ) ( ) cosA f t t dtα α

π�

�� (7.21)

1( ) ( ) sinB f t t dtα α

π�

�� (7.22)

the above equation can also be written as

0( ) [ ( ) cos ( ) sin ]f x A x B x dt dα α α α α

�� (7.23)



�� "��#�� $��#��#

If ( ) ( ), i.e., if ( )f x f x f x� � � is an odd function, Eq. (7.21) gives

1( ) ( ) cosA f x x dxα α

π�

��

If ( )f x is an odd function, then

1 1( ) ( ) cos ( ) cos

1( ) cos ( )

A f x x dx f x x dx

f x x dx A

α α απ π

α απ

��

� �

�

��

� � � � �

� � � �

� �

�implying 2 ( ) 0 or ( ) 0,A A� �� i.e.,

1( ) ( ) cos 0A f x x dxα α

π�

�� (7.24)

Also,

0

1( ) ( ) sin

1( ) sin if ( ) is an odd function

1 2( ) sin ( ) sin

B f x x dx

f x x dx f x

f x x dx f x x dx

α απ

απ

α απ π

�

��

��

�

� �

��

�

� �

� �

�

�

� � (7.25)

Thus, Eq. (7.23) reduces to

0( ) ( ) sinf x B x dα α α

�� (7.26)

which is the Fourier sine integral representation, where ( ) and ( )A B� � are defined by the

relations (7.24) and (7.25). Similarly, if ( )f x is an even function, i.e., if ( ) ( ),f x f x� � then

we obtain the Fourier cosine integral representation

0( ) ( ) cosf x A x dα α α

�� (7.27)

where

0

2( ) 0, ( ) ( ) cosB A f x x dxα α α

π�

� � � (7.28)



��% ��&��

From the Fourier Integral Theorem 7.1, we have

0

1( ) ( ) cos ( )f x f t t x dt dα α

π� �

�� (7.29)

In terms of the complex exponential function, Eq. (7.29) takes the form

( ) ( )

0

( ) ( )

0 0

1( ) ( ) [ ]

2

1( ) ( )

2

i t x i t x

i t x i t x

f x f t e e dt d

f t e dt d f t e dt d

α α

α α

απ

α απ

� � � � ��

� � � ��

� �

� ��

� �

� � � �Let � �� in the second integral; then it becomes

0( ) ( )

0( ) ( )i t x i t xf t e dt d f t e dt dα αα α

��

� �� Hence,

( )1( ) ( )

2i t xf x f t e dt dα α

π� � ��

� � � (7.30)

This is the exponential form of the Fourier integral theorem. Equation (7.30) can be rewrittenas

1 1( ) ( )

2 2i t i xf x f t e dt e dα α α

π π� � ��

� �� (7.31)

Thus, we formally define the Fourier transform pair as follows:

Definition 7.2 Let ( )f x be a function defined on ( , )�� and is piecewise continuous,

differentiable in each finite interval and is absolutely integrable on ( , ).�� From Eq. (7.31), if

1( ) ( )

2i tF f t e dt�

�

�

�

��

� � (7.32)

then we have, for all x,

1( ) ( )

2i xf x F e dαα α

π� ��

� � (7.33)

Here, ( )F � defined by Eq. (7.32) is the Fourier transform of ( ),f x and ( )f x defined

by Eq. (7.33) is called the Inverse Fourier transform of ( )F � and is denoted by



( ) [ ( ); ]F f t� �� (7.34)

1( ) [ ( ); ]f x F x��

� � (7.35)

which constitute the Fourier transform pair.

We have seen in Section 7.2.2 that if ( )f x is an odd function, the Fourier integral

representation of ( )f x reduces to

0( ) ( ) sinf x B x dα α α

��

or

0 0

2( ) sin ( ) sinf x x f t t dt dα α α

π� �

� � � (7.36)

If

0

2( ) ( ) sin [ ( ); ]

�� s sF f t t dt f t� � �

�� (7.37)

then

1

0

2( ) ( ) sin [ ( ); ]

� �� s s sf x F x d F x� � � �

�� (7.38)

Here, Eq. (7.37) is the Fourier sine transform of ( )f x and its inverse sine transform is given

by Eq. (7.38).

Similarly, when ( )f x is an even function, we can obtain the Fourier cosine transform and

the corresponding inverse as

0

2( ) ( ) cos [ ( ); ]

�� c cF f t t dt f t� � �

�� (7.39)

1

0

2( ) ( ) cos [ ( ); ]

� �� c c cf x F x d F x� � � �

�� (7.40)

��' ��&��&��(��

EXAMPLE 7.1 Find the Fourier transform of

2/2( ) xf x e−=



Solution Following the definition of Fourier transform, we have

2

2 2

/2

( ) /2 /2

1[ ( ); ] ( )

2

1

2

1

2

�

��

� ��

� � � ��

�

�

�

i x

x i x

x i

f x f x e dx

e e dx

e e dx

�

�

� �

�

�

�

�

�

�

�

�

Let

2

x it

��

�

Then

2

dxdt�

Thus

22/2

[ ( ); ]� � �

�� te

f x e dt�

�

��

or

22/2/2( )

eF e

ααα π

π

−−= =


| |( ) ,a xf x e x��

Solution We know that

, 0| |

, 0

x xx

x x

� ��

��

Therefore,

0

0

1[ ( ); ] ( )

2

1 1

2 2

i x

ax i x ax i x

F f x f x e dx

e e dx e e dx

α

α α

απ

π π

�

��

� ��

�

� �

�

� �



0 ( ) ( )

0

2 2

1 1

2 2

1 1 1 2

2

a i x a i xe dx e dx

a

a i a i a

α απ π

π α α π α

��

� �

� � � ��

� �

EXAMPLE 7.3 Find the Fourier transform of ( )f x defined by

1, | |( )

0, | |

x af x

x a

��

��

and hence evaluate

0

sin cos sin,

a x ad d

α α αα αα α

� �

�� Solution From the definition of the Fourier transform,

1( ) [ ( ); ] ( )

2

1 1

2 2

1 2

22 2

i x

ai xa i x

a a

i a i a i a i a

F f x f x e dx

ee dx

i

e e e e

i i i

�

��

� � � �

� ��

��

� ��

�

��

� �

� �

� �

� ��

� � � ��

�

�

�

Therefore,

0

2 sin, 0

2( )

2 sin 2Lt , 0

2 2�

��

� ��

��

� � ��

��

� ��

a

Fa a a

a

Now,

1 1( ) [ ( ); ] ( )

2

��

� � i xf x F x F e d��

��

Thus,

1, | |1 2 sin( )

0, | |2 2i x x aa

f x e dx a

αα απ α π

� ��

��



i.e.,

1, | |1 sin (cos sin )

0, | |

x aa x i xdx

x a

α α απ α

�

��

��

Hence,

, | |sin cos

0, | |

x aa xd

x a

πα α αα

�

��

��

Also, by setting 0x � in the above equations, we obtain

sin ad

α α πα

�

��

Since the integrand is even, we can have

0

sin

2

ad

α παα

��

EXAMPLE 7.4 Find the Fourier cosine and sine transforms of bxe� and evaluate the integrals

(i) 2 20

cos xd

b+

� ��

�

�(ii)

2 20

sin xd

b

α α αα +

�

�

Solution Given ( ) bxf x e�� and following the definitions of Fourier cosine and sine

transforms, viz.

0

2( ) [ ( ); ] ( ) cosc cF f x f x x dx� � �

�

��

� � � � �0

2( ) [ ( ); ] ( ) sins sF f x f x x dx� � �

�

��

we obtain

0

2[ ; ] cosbx bx

c e e x dx� �

�

�� (7.41)

0

2[ ; ] sinbx bx

s e e x dx� �

�

�� (7.42)



Let

10

cosbxI e x dxα� ��

20

sinbxI e x dxα� ��

Integrating 1I by parts, we have

1 200

1 1cos sinbx bxI e x e x dx I

b b b b

α αα α� �� (7.43)

Integrating 2I by parts, we have

2 100

1sin cosbx bxI e x e x dx I

b b b

α αα α� �� (7.44)

Solving Eqs. (7.43) and (7.44) for 1 2andI I , we obtain

1 22 2 2 2,

bI I

b b

�

� �

� �

� �

Hence,

2 2 2 2

2 2( ) , ( )c s

bF F

b b

��

� ��

� �

� �

Then,

0

2( ) ( ) coscf x F x dα α α

π�

� �i.e.

2 20

2 2cosbx b

e x db

α απ π α

��

or

2 20

cos

2bxx

ebb

α πα

� ��


2 20

sin

2bxx

eb

α α πα

� ��



EXAMPLE 7.5 Find the Fourier sine transform of ( ),f x if

0, 0

( ) ,

0,

x a

f x x a x b

x b

� ��

� � ��

Solution Following the definition of the Fourier sine transform, we have

0

2

2( ) ( ) sin

2sin

2 cos 1cos

2 cos cos sin sin

s

b

a

b b

aa

F f x x dx

x x dx

x xx dx

a a b b b a

α απ

απ

α απ α α

α α α απ α α

��

�

� ��

� ��

�

�

�

��) �� &

In many practical situations, determination of Fourier transform of certain functions is verycomplex. Once we know the transform of some elementary functions, we can find the transformof many other functions with the help of the properties associated with the Fourier transform.We now discuss some of the important properties of the Fourier transform.

Theorem 7.2 (Linearity property). If ( ) and ( )F G� � are the Fourier transforms of

( ) and ( )f x g x respectively, then

1 2 1 2[ ( ) ( ); ] ( ) ( )c f x c g x c F c G� � ��

11 2 1 2[ ( ) ( ); ] ( ) ( )c F c G x c f x c g x� �

�

� � ��

where 1 2andc c are constants.

Proof

1 2 1 2

1 2

1 2

1[ ( ) ( ); ] [ ( ) ( )]

2

( ) ( )2 2

( ) ( )

i x

i x i x

c f x c g x e c f x c g x dx

c ce f x dx e g x dx

c F c G

�

� �

��

� �

� �

�

��

� �

��

� � �

� �

� �

�

� �

�



Theorem 7.3 (Change of scale). If ( )F � is the Fourier transform of ( ),f x then the Fourier

transform of ( )f ax is

1F

a a

��

Proof From the definition of the Fourier transform, we have

1( ) [ ( ); ] ( )

2

�

�� i xF f x e f x dx�

� �

��

Hence,

1[ ( ); ] ( )

2i xf ax e f ax dx�

�

�

�

��

Setting ,ax t� we have / .�dx dt a Therefore,

1 1[ ( ); ] exp ( )

2f ax i t f t dt

a a

��

�

��

� ��

Hence,

1[ ( ); ] ( / ), 0f ax F a a

a� �� (7.45)


1[ ( ); ] ( / ), 0f ax F a a

a� �� (7.46)

Combining Eqs. (7.45) and (7.46), we have the property

1[ ( ); ] ( / ), 0

| |f ax F a a

a� �� (7.47)

It can also be established that

1[ ( ); ] ( / ), 0s sf ax F a a

a� �� (7.48)

1[ ( ); ] ( / ), 0c cf ax F a a

a� �� (7.49)

Theorem 7.4 (Shifting property). If ( )F � is the Fourier transform of ( ),f x then the Fourier

transform of ( ),f x a� i.e.,

[ ( ); ] ( )i af x a e F��



Proof From the definition of Fourier transform

1( ) [ ( ); ] ( )

2i xF f x e f x dx��

�

�

��

Therefore,

1[ ( ); ] ( )

2i xf x a e f x a dx��

�

�

��

Setting

,� �x a t

we have .dx dt� Then

( )1[ ( ); ] ( )

2i a tf x a e f t dt��

�

� ��

� � ��

Hence,

[ ( ); ] ( )i af x a e F�� (7.50)


1[ ( ); ] ( )i ae F x f x a� �� (7.51)

Theorem 7.5 (Modulation property). If ( )F � is the Fourier transform of ( ),f x then the

Fourier transform of ( ) cosf x ax is

1[ ( ) ( )]

2F a F a� ��

Proof From the definition of Fourier transform, we have

1( ) [ ( ); ] ( )

2i xF f x e f x dx��

�

�

��

Therefore,

( ) ( )

1[ ( ) cos ; ] ( )

22

1 1 1( ) ( )

2 2 2

1[ ( ) ( )]

2

i x i xi x

i a x i a x

e ef x ax e f x dx

e f x dx e f x dx

F a F a

� ��

� �

��

� �

� �

��

��

� ��

� ��

��

� � � �

�

� �

�

��



In the same fashion, it can be established that

� �

0

0

0 0

2[ ( ) cos ; ] ( ) cos sin

2 1( ) [sin ( ) sin ( ) ]

2

1 2( ) sin ( ) ( ) sin ( )

2

1[ ( ) ( )]

2

s

s s

f x ax f x ax x dx

f x a x a x dx

f x a x dx f x a x dx

F a F a

� ��

� ��

� ��

� �

�

�

� �

�

� � � �

� � � � � �

�

� � � �

�

�

� �

�

� ��

1

[ ( ) cos ; ] [ ( ) ( )]2c c cf x ax F a F a� � �� (7.54)

1

[ ( ) sin ; ] [ ( ) ( )]2s c cf x ax F a F a� � �� (7.55)

1

[ ( ) sin ; ] [ ( ) ( )]2c s sf x ax F a F a� � �� (7.56)

Theorem 7.6 (Differentiation). If ( )f x and its first ( 1)r � derivatives are continuous, and

if its rth derivative is piecewise continuous, then

( )[ ( ); ] ( ) [ ( ); ], 0,1, 2, ...r rf x i f x r� � ��

provided f and its derivatives are absolutely integrable. In addition, we assume that ( )f x and

its first ( 1)r � derivatives vanish as .x ��

Proof From the definition, we have the Fourier transform of /r rd f dx as

( ) ( )1[ ( ); ] ( ) (say)

2

rr i x r

r

d ff x e dx F

dx��

�

�

��


1 1

1 1

1 1 1( )

2 2 2

r r ri x i x i x

r r r

d f d f d fe dx e i e dx

dx dx dxα α αα

π π π

��

� ��

� ��

If we assume that 1 1/r rd dx− − tends to zero as ,x �� we may write the above result in the

form( ) ( 1) 2 ( 2)( ) ( ) ( ) ( ) ( ) ( ) ( )� � � � � � �

� �

� � � � � ��r r r rF i F i F i F



Hence,( ) ( ) ( ) ( )� � ��

r rF i F

and, therefore,

( )[ ( ); ] ( ) ( )r rf x i F� � �� (7.57)

EXAMPLE 7.6 Let ( ) ( )rcF � be the Fourier cosine transform of /r rd f dx and ( ) ( )�r

sF be

the Fourier sine transform of / .r rd f dx Then prove that

1(2 ) 2 2

2 2 10

1(2 1) 2 2 1

2 20

( ) ( 1) ( 1) ( )

( ) ( 1) ( 1) ( )

rr n n r r

c r n cn

rr n n r r

c r n sn

F a F

F a F

α α α α

α α α α

−

− −=

−+ +

−=

= − − + −

= − − + −

∑

∑assuming

1

1Lt 0

r

rx

d f

dx

�

��

� ��

1

110

2Lt (say)

r

rrx

d fa

dxπ

−

−−→=

Solution From the definition, we have

( )

0

2( ) cos

rr

c r

d fF x dx

dxα α

π�

� � (7.58)

( )

0

2( ) sin

rr

s r

d fF x dx

dxα α

π�

� � (7.59)

Integrating Eq. (7.58) by parts, we get

1 1( )

1 100

2( ) cos ( ) sin ( )

r rr

c r r

d f dF x x dx

dx dxα α α α

π

��

� �

� � ��

� ��

Now, we assume that

1

1Lt 0

r

rx

d f

dx

�

��

� ��



1

110

2Lt (say)

r

rrx

d fa

dxπ

−

−−→=

Then

( ) ( 1)1( ) ( )r r

c r sF a F� � ��

�

� � � (7.60)

Integrating Eq. (7.59) by parts, we have

1 1( ) 1

1 100

2( ) sin cos ( ) ( )

r rr r

s cr r

d f dF x x dx F

dx dxα α α α α α

π

��

� �� (7.61)


( ) 2 ( 2)1( ) ( )r r

c r cF a Fα α α−−= − −

By repeated application of these results, we can show ( ) ( )rcF � to be a sum of a’s and either

(1) ( )cF � or ( ),cF � depending on whether r is odd or even. (1)cF will be present if r is odd

and may be replaced by 0 ( ).sa F� �� Similar arguments give us the formulae

1(2 ) 2 1 2

2 2 10

( ) ( 1) ( 1) ( )� � � �

�

�

� �

�

� � � � ��r

r n n r rc r n c

n

F a F

1(2 1) 2 1 2 1

2 20

( ) ( 1) ( 1) ( )� � � �

�

� � �

�

�

� � � � ��r

r n n r rc r n s

n

F a F

Note: When 0x � and 3

30,

df d f

dx dx� �

22

20

44

40

2cos ( )

2cos ( )

c

c

d fx dx F

dx

d fx dx F

dx

α α απ

α α απ

�

�

� �

�

�

�

Similarly, when 2

20 and 0,

d fx f

dx� � �

22

20

44

40

2sin ( )

2sin ( )

s

s

d fx dx F

dx

d fx dx F

dx

α α απ

α α απ

�

�

� �

�

�

�



EXAMPLE 7.7 Find the Fourier cosine transform of 2exp ( ).at�

Solution We have from the definition of Fourier cosine transform,

2 2

0

2[ ; ] cos (say)at at

c e e t dt I� ��

�� (7.62)

Differentiating with respect to ,� we obtain

2 2

2 2

0 0

0 0

2 1 2sin sin ( )

2

1 2[( sin )] cos

2

at at

at at

dIte t dt t d e

d a

e t e t dta

α αα π π

α α απ

� ��

��

� � �

� ��

� �

�Therefore,

2

dII

d a

αα

= −

i.e.,

2

dId

I a

��


2/ 4��

aI ce (7.63)

But when 0� � , from Eq. (7.62) we have

2

0

2 atI e dtπ

� �� From Example 7.1, using change of scale property, we obtain

2 1 1

2 2I

a a

�

�

� �

From Eq. (7.63) we get 1

.2

ca

� Hence,

2 2 / 41[ ; ]

2at a

c e ea

��



EXAMPLE 7.8 If the Fourier sine transform of 2( ) is /(1 ),� ��f x find ( )f x .

Solution From the definition, we have

20

2

20

20 0

2( ) sin

1

2 ( 1) 1sin

(1 )

2 sin 2 sin

(1 )

f x x d

x d

x xd d

α α απ α

α α απ α α

α αα απ α π α α

∞

∞

∞ ∞

=+

+ −=+

−+

�

�

�

� �But

0

sin

2

xd

α παα

��

Hence,

20

2 sin( )

2 (1 )

xf x d

π α απ α α

��

�� (7.64)

20

2 cos

1

df xd

dx

α απ α

��

�� (7.65)

2

2 20

2 sin

1

d f xd

dx

α α απ α

��

�� (7.66)

From Eqs. (7.64) and (7.66), it follows that

2

2 0

2 sin0

2

d f xf d

dx

α παπ α

��

whose solution is found to be

1 2x xf c e c e�� (7.67)

Therefore,

1 2x xdf

c e c edx

−= −

When, 0,x � from Eq. (7.64) we have (0) /2,��f and from Eq. (7.65),

20

(0) 2

21

df d

dx

α ππ α

��

��



Also, from Eq. (7.67), using these results, we get

1 2 1 2,2 2

c c c cπ π+ = − = −

Solving the above two equations, we get 1 20, /2.�� c c Thus, ( ) /2 .��

�

xf x e

EXAMPLE 7.9 If the Fourier cosine transform of ( )f x is ,n ae ��

� find ( ).f x

Solution Using the definition, we have

0

2( ) cosn af x e x dαα α α

π� �� (7.68)

But we know from calculus that

2 20cosa a

e x da x

α α α� � �

�� (7.69)

Differentiating this result n times with respect to a, we get

2 20

1 1

1 1

( 1) cos

1 1 1

2

1[( 1) ! ( ) ( 1) ! ( ) ]

2

( 1) ![( ) ( ) ]

2

nn n a

n

n

n

n n n n

nn n

d ae x d

da a x

d

a ix a ixda

n a ix n a ix

na ix a ix

αα α α� �

� � � �

� � � �

� ��

� ��

� � � � � �

��

�

Let

(cos sin )a ix r i� ��

Then

cos , sina r x r� ��

Therefore,

2 2 2 , tanx

r a xa

��

Thus,

1 1( ) [cos ( 1) sin ( 1) ]n na ix r n i n� ��

� � � � � � �

1 1( ) [cos ( 1) sin ( 1) ]n na ix r n i n� ��

� � � � � � �



Then

2 2 ( 1) 20

!cos ( 1)cos

( )n a

n

n ne x d

a xα θα α α

� ��

��

Hence from Eq. (7.68),

2 2 ( 1/2)

2 !cos ( 1)( )

( )

�

� �

��

�n

n nf x

a x


(i) /n nu x� � of the function ( , )u x t assuming that u and its first (n – 1) derivatives with

respect to x vanish as .x ��

(ii) / .u t� �

Also, find the sine and cosine Fourier transforms of 2 2/u x� � of the function ( , )u x t .

Solution (a) We shall adopt the following notation: The Fourier transform of ( , )u x t

with respect to the variable x is defined as

1[ ( , ); ] ( , ) ( , )

2i xu x t x e u x t dx U t��

�

�

�� (7.70)

Then the Fourier transform of /u x� � is

1( , );

2i xu u

x t x e dxx x

��

�

��

� ��

�

Integration by parts yields

1[ ( , ) ] ( , )

2i x i xu x t e i u x t e dxα αα

π��

��

� ��

If we assume

Lt ( , ) 0x

u x t� �

�

then we find that

( , ); [ ( , ); ] ( , )u

x t x i u x t x i U tx

� � � � ��

� � (7.71)



Similarly, the Fourier transform of 2 2/u x� � is

2 2

2 2

2

( , ) 1 1;

2 2

1( ) ( )

2

i x i x

i x i x i x

u x t u ux e dx e d

xx x

ue i e u i e u dx

x

� �

� � �

��

� ��

� �

��

��

��

� � ��

��

� �

�

� � ��

��

�

Assuming that both u and /u t� � tend to zero as ,x �� we have

22 2 2 2

2( , ); ( 1) ( ) [ ( , ); ] ( 1) ( ) ( , )

ux t x i u x t x i U t

x� � � � �

� ��

��

� � (7.72)

Thus, in general, the Fourier transform of the nth derivative of ,( )tu x is given by

( , ); ( 1) ( ) [ ( , ); ] ( 1) ( ) ( , )

nn n n n

n

u x tx i u x t x i U t

x� � � � �

� ��

��

� � (7.73)

(b) Now

, ,1 1

( , ); ( ) ( ) ( , )2 2

i x i xtt t

u ux t x e x dx e u x dx U t

t t t� ��

� �

� �

��

� � � � ��

�

Therefore,

( , ); ( , )tu

x t x U tt

� ��

� (7.74)

(c) In the case of Fourier sine and cosine transforms, we have

2 2

2 20

00

2( , ); sin

2 2sin cos

su u

x t x x dxx x

u ux x dx

x x

� ��

� � ��

�

��

� ��

� ��

�

�

� ��

� ��

�

We assume that 0 as .u

xx� � ��

�Then the RHS term of the above equation becomes

00 0

2 2cos [ ( , ) cos ] ( , )sin

ux dx u x t x u x t x dx

xα α α α α α

π π� ��

� ��



Also, assuming that ( , ) 0 as ,u x t x� �� this equation becomes

2

0

2( , ) [ ( , ); ]s

xtu x t u x x� � �

��

� ��

Hence,

� � � ��

� ��

22

20

2( , ); ( , ) [ ( , ); ]s s

x

tu

x t x u x t u x xx

��

� � (7.75)

Similarly, it can be shown that if

( , ) 0 and 0 asu

u x t xx

� � � ��

then

22

20

2 ( , )( , ); [ ( , ); ]c c

x

tu u x t

x t x u x xxx

� � ��

� ��

� �

�� (7.76)

Obviously, the choice of the sine or cosine transform is dccided by the form of the boundarycondition at the lower limit of the variable selected for exclusion. Thus, we observe that for

the exclusion of 2 2/u x� � from a given PDE, we require

0

0

in the case of sine transform

in the case of cosine transform

x

x

u

u

x

�

�

��

��* ��+��,��&� -��,��&.

If ( ) and ( )F G� � are the Fourier transforms of the functions ( ) and ( ),f x g x then the product

( ) ( )F G� � is the Fourier transform of the convolution product * .f g

Proof The product

1* ( ) ( )

2f g f u g x u du

π�

�� (7.77)

is called the convolution or Faltung of the functions andf g over the interval ( , ).�� Then

the Fourier transform of this convolution integral yields

1[( * ); ] ( ) ( )

2

1( ) ( )

2

i x

i x

f g e f u g x u du dx

e f u g x u du dx

�

�

��

�

� �

��

� �

��

� �

� �

� �

� �

�

(7.78)



Since f and g are absolutely integrable, the order of integration can be interchanged and,therefore,

( )1[( * ); ] ( ) ( )

2i x u i uf g f u g x u e e dx du� ��

��

� ��

Let .x u y� � Then .dx dy� Therefore,

1[( * ); ] ( ) ( )

2

1 1( ) ( )

2 2

( ) ( )

i u i y

i u i y

f g f u e g y e dy du

e f u du e g y dy

F G

� �

� �

��

� �

� �

� �

��

� �

��

��

�

�

� �

� �

�

(7.79)

Hence the theorem is proved.We can verify that

* * ,�f g g f (7.80)

i.e.

1* ( ) ( )

2f g f u g x u du

π�

�� (7.81)

Setting ,x u �� we have .du d�� Then

1 1* ( ) ( ) ( ) ( ) *

2 2f g f x g d g u f x u du g fα α α

π π� �

�� (7.82)

Hence, the convolution is commutative.

Special cases (sine and cosine convolution integrals)(i)

0 0 0

0 0

0 0

2( ) ( )cos ( ) cos ( ) cos

2( ) ( ) cos cos

1( ) ( ) [cos | | cos ( ) ]

2

c c c

c

c

F G x d F x d g d

g d F x d

g d F x x d

α α α α α α α η ηα ηπ

η η α α ηα απ

η η α η α η α απ

� � �

� �

� �

�

�

� � � �

� � �

� �

� �



But,

0 0

2 1( ) ( )cos ( ) [ (| |) ( )]

2cf x F x d g d f x f xα α α η η η ηπ

� �� (7.83)

(ii) If ( ) and ( )s sF G� � are the Fourier sine transforms of ( ) and ( )f x g x , then we can

show that

0 0

1( ) ( )sin ( ) [ (| |) ( )]

2s sF G x d f g x g x dα α α α η η η η� �

� � � �� (7.84)

(iii) 0

( ) ( ) sinc sF G x d� � � ��

�

0 0

0 0

0 0

0

2( ) sin ( ) sin

2( ) ( ) sin sin

1( ) ( ) [cos | | cos ( ) ]

2

1( ) [ (| |) ( )]

2

c

c

c

F x d g d

g d F x d

g d F x x d

g f x f x d

� � � � ��

� � � � ��

� � � � � � � ��

� � � �

� �

� �

� �

�

�

�

� � � �

� � � �

� �

� �

� �

� (7.85)

(iv)0 0

1( ) ( )sin ( ) [ (| |) ( )]

2s cF G x d f g x g x dα α α α η η η η� �

� � � �� (7.86)

EXAMPLE 7.11 If ( ) and ( )c cF G� � are the Fourier cosine transforms of ( ) and ( )f x g x

respectively, show that ( ) ( )c cF G� � is the transform of

0

1( ) [ (| |) ( )]

2f u g x u g x u du

��

Solution Refer Eq. (7.83).

�� +��/��

From the Fourier convolution theorem, we have

0

1 1( ) ( ) * ( ) ( )

2 2i xe F G d f g f u g x u duα α α α

π π� ��

� � ��



If we set 0x � , the above equation reduces to

1 1( ) ( ) ( ) ( )

2 2F G d f u g u duα α α

π π� �

��

For the case ( ) *( ),� �g u f u where *( )f u is the complex conjugate of the function ( ),f u we have

( ) [ ( ); ] [ *( ); ] *( )G g u f u F� � � �� Thus,

1 1( ) *( ) ( ) * ( )

2 2F F d f u f u duα α α

π π� �

�� (7.87)

Therefore,

2 2| ( ) | | ( ) |F d f u duα α� �

�� (7.88)

Equation (7.88) is known as Parseval’s relation.

EXAMPLE 7.12 Using the Fourier cosine transform of and ,ax bxe e� � show that

2 2 2 20, 0, 0

2 ( )( ) ( )

da b

ab a ba b

α πα α

��

�� (7.89)

Solution Let ( ) , ( ) ;bx axf x e g x e� �

� � then

2 20 0

2 2 2( ) ( )cos cosbx

cb

F f x x dx e x dxb

α α απ π π α

� � ��


2 2

2( )c

aG

a�

� �

�

�

However,

0 0 0

0 0

0

2( ) ( ) ( ) ( ) cos

2( ) ( ) cos

( ) ( )

c c c

c

F G d F d g x x dx

g x dx F x d

f x g x dx

α α α α α απ

α α απ

� � �

� �

�

�

�

�

� � �

� �

�i.e.

0 0( ) ( ) ( ) ( )c cF G d f x g x dxα α α

� ��



Hence, it follows that

( )

0 0

1( ) ( ) a b x

c cF G d e dxa b

α α α� � � ��

�� Therefore,

2 2 2 20, 0, 0

2 ( )( ) ( )

da b

ab a ba b

α πα α

��

��

��0 ��&��

The Dirac delta function has been defined in Chapter 3. The details of its properties can befound in Section 3.4. We may recall the shifting property of the delta function, i.e.

( ) ( ) ( )t a f t dt f aδ�

��

and then obtain its Fourier transform as

1[ ( ); ] ( ) / 2

2i t i at a e t a dt e� ��

�

�

�� (7.90)

When 0,a � we obtain the formal result

1[ ( ); ]

2t� �

�� (7.91)

That is, the Fourier transform of the Dirac delta function ( )t� is constant and equal to 1/ 2 .πIt then follow that

1 1[1; ] 2 ( )

2i tt e d t� � ��

�

��

� �� (7.92)

��1 &�� &�

As already discussed, the idea of a Fourier transform and its inverse can be extended tofunctions involving two or more variables.

Let ( , )f x y be a function of two variables x and y. Let it be piecewise continuous and

satisfy the condition

| ( , ) |f x y dx dy� �

�� (7.93)

Then the Fourier transform pair can be written as

( )1( , ) ( , )

2i x yF e f x y dx dyα βα β

π� � ��

� � � (7.94)



( )1( , ) ( , )

2i x yf x y e F d dα β α β α β

π� � � ��

� � � (7.95)

We can split Eq. (7.94) into two steps: first by treating ( , )f x y as a function of x and then

treating the result as a function of y sequentially, i.e.,

1*( , ) ( , )

2

1( , ) *( , )

2

i x

i y

f y e f x y dx

F e f y dy

α

β

απ

α β απ

�

��

�

��

�

�

�

�Similarly, the inversion formula (7.95), can be written as

1( , ) *( , )

2

1*( , ) ( , )

2

i x

i y

f x y e f y d

f y e F d

α

β

α απ

α α β βπ

� ��

� ��

�

�

�

�Assuming that the partial derivatives of f occurring in the equation are absolutely integrable

and that , / , /f f x f y� � � � tend to zero at infinity, the double Fourier transform of derivatives

yield the following results:

( , ); , ( , )f

x y x y i Fx

� � � � ��

� (7.96)

� � � �2

( , ); , ( , )f

x y x y Fx y

� � ��

� � � ��

� ��

The convolution property leads to the following results:

1 1[ ( , ) ( , ); , ] ( ) ( ) ( , )

2F G x y f x u g x u g u v du dv� � � � � �

��

� � � � ��

The Fourier transform in the case of three variables is

( )3/2

1( , , ) ( , , )

(2 )i x y zF e f x y z dx dy dzα β γα β γ

π

� � � � ��

� � � � (7.99)

��2 ��&

The Fourier transform technique outlined so far is applicable to problems involving infiniteor semi-infinite domains. But, in many practical situations, we come across finite intervals inboundary value problems. Therefore, it is natural to extend Fourier transform method toproblems in which the range of an independent variable is finite. It is then possible to find



their inverses from the well-known theory of Fourier series. It may be recalled that if a

function ( )f x satisfies Dirichlet conditions in the interval 0 ,x �� then it has Fourier sine

series

1

( ) sinnn

f x b nx�

��

in which

0

2( )sin , 1, 2, 3,

�

�� nb f x nx dx n (7.101)

The Fourier series in Eq. (7.100) converges pointwise to ( )f x at points where ( )f x is

continuous and to the value (1/2)[ ( ) ( )]f x f x+ + − at other points. Similarly, ( )f x has Fourier

cosine series

0

1

( ) cos2 n

n

af x a nx

�

�� (7.102)

in which

0

2( )cos , 0,1, 2, ...

�

�

� ��na f x nx dx n ��

��2�� #3��!

If ( )f x satisfies Dirichlet conditions in the interval 0 ,x �� then we define finite sine

transform of ( )f x by

0[ ( ); ] ( ) ( )sin , 1, 2, 3, ...s sf x n F n f x nx dx n

�� (7.104)

which is a sequence of numbers. Comparing Eqs. (7.101) and (7.104), we notice that

2( ), 1, 2, 3, ...n sb F n n

�

� � (7.105)

Now from Eq. (7.100) we can have the result

1

1

2[ ( ); ] ( ) ( ) sins s s

n

F n x f x F n nx�

��

��

which is the inverse finite sine transform. Similarly, we can define finite sine transform whenthe independent variable x lies in the interval (0, L),

0[ ( ); ] ( ) ( )sin , 1, 2, 3,

L

s sn x

f x n F n f x dx nL

�� (7.107)



with

1

1

2[ ( ); ] ( ) ( ) sin , 0s s s

n

n xF n x f x F n x L

L L

��

�� (7.108)

as the corresponding inversion formula.

��2� ��#��#3��!

If ( )f x satisfies the Dirichlet conditions in the interval 0 ,x �� we define the finite cosine

transform of ( )f x as

0[ ( ); ] ( ) ( )cos , 0,1, 2, ...c cf x n F n f x nx dx n

�� (7.109)

The corresponding inversion formula is

1

1

(0) 2[ ( ); ] ( ) ( ) cosc

c c cn

FF n x f x F n nx

� �

��

�� (7.110)

which holds at each point in the interval (0, )� at which ( )f x is continuous. Similarly, when

the independent variable x lies in the interval (0, )L , the corresponding pair assumes the form

0[ ( ); ] ( ) ( )cos , 0,1, 2, ...

L

c cn x

f x n F n f x dx nL

�� (7.111)

1

1

(0) 2[ ( ); ] ( ) ( ) cosc

c c cn

F n xF n x f x F n

L L

��

��

�� (7.112)

Having defined the finite cosine transform, we shall attempt to find some results involvingderivatives up to Fourth order to facilitate the solution of a few boundary value problems,which are actually presented under miscellaneous examples. For instance, if f is a function

of x and t, 0 , 0,x L t� � � then we have

0 00; sin ( , )sin cos

LL L

sf f n x n x n n x

n dx f x t f dxx x L L L L

� � � ��

�

Therefore,

; [ ; ] ( )s c cf n n

n f n F nx L L

� ��

� � (7.113)

0 00; cos ( , ) cos sin

[ ; ] { (0, ) ( , ) cos }

LL L

c

s

f f n x n x n n xn dx f x t f dx

x x L L L L

nf n f t f L t n

L

� � � �

� �

� � � ��

�

� ��

�

� ��



Considering second order derivatives, we obtain

2 2

2 20 00; sin sin cos

LL L

sf f n x f n x n f n x

n dx dxL x L L x Lx x

� � � ��

� ��

00cos sin

[ ; ] [ (0, ) ( , ) cos ]

LL

s

n n x n n xf f dx

L L L L

n nf n f t f L t n

L L

� � � �

� � �

� ��

� ��

�

�

Therefore,

2 2 2

2 2; [ ; ] { (0, ) ( , ) cos }s s

f n nn f n f t f L t n

Lx L

� � ��

� � � ��

� � ��

In particular, if (0, ) ( , ) 0,f t f L t� � this result simplifies to

2 2 2

2 2; [ ; ]s s

f nn f n

x L

��

��

� � (7.116)


2 2 2

2 2; [ ; ] { (0, ) ( , ) cos }c c x x

f nn f n f t f L t n

x L

� ��

� � � ��

� � (7.117)

In case /f x� � vanishes at the ends 0 and ,x x L� � it simplifies to

2 2 2

2 2; [ ; ]c c

f nn f n

x L

��

��

� � (7.118)

By repeatedly applying these results, we can deduce that

4 4 4

4 4; [ ; ],s s

f nn f n

x L

��

��

� � (7.119)

if andx xxf f vanish at both the ends 0, ,x x L� � and

4 4 4

4 4; [ ; ]c c

f nn f n

x L

��

� ��

� � (7.120)



provided

0 when 0;x xxxf f x x L� � � �

�� 4��

Let us consider the problem of flow of heat in an infinite medium ,x�� when the

initial temperature distribution ( )f x is known and no heat sources are present. Mathematically,

we have to solve the problem described in the following example.

EXAMPLE 7.13 Solve the heat conduction equation given by

2

2PDE: , , 0

u uk x t

tx� � � � � � ��

subject to

BCs: ( , ) and ( , ) both 0 as | |

IC: ( , 0) ( ),xu x t u x t x

u x f x x

� ��

Solution Taking the Fourier transform* of PDE, we get

2 [ ( , ); ] [ ( , ); ]tk u x t x u x t x� � �� or

2( , ) ( , ) 0tU t k U t� � �� (7.121)

In deriving this, the BCs are already utilized (as can be seen from Eqs. (7.72) and (7.74). TheFourier transform of the IC gives

( , 0) ( ), � � � U Fα α α� (7.122)

The solution of Eq. (7.121) can be readily seen to be

2k tU Ae ��

�

When 0,t � we have from Eq. (7.122) the relation ( ),U F �� implying ( ).A F �� Therefore,

2( , ) ( ) k tU t F e ��

�

�(7.123)

Inverting this relation, we obtain

2 21 1( , ) ( ) ( )exp ( )

2 2k t i xu x t F e e d F k t i x dα αα α α α α α

π π� ��

� � � ��

*When the range of spatial variable is infinite, the Fourier exponential transform is used rather than thesine or cosine transform.



The product form of the integrand in Eq. (7.124) suggests the use of convolution. If the

Fourier transform of ( )g x is 2

,k te �� then ( )g x will be given by

21( )

2k t i xg x e e dα α α

π� � ��

� �But, if 0,a b� is real or complex, and we know that

2 2exp ( 2 ) exp ( / )ax bx dx b aa

π�

�� (7.125)

Here, , 2 .a kt b ix� � Therefore,

2 21 1( ) exp exp

2 4 2 4

�

�

� � � ��

� � � �x x

g xkt kt kt kt

Using the convolution theorem, we have

1( , ) ( ) ( )

2

�

�� u x t f g x dα α α

π � (7.126)

Hence,

2

2

1 1 ( )( , ) ( ) exp

2 2 4

1 ( )( ) exp

4 4

xu x t f d

kt kt

xf d

kt kt

αα απ

αα απ

�

��

�

��

��

��

�

� (7.127)

Introducing the change of variable

4

� �

�

xZ

kt

we can rewrite solution (7.127) in the form

21( , ) ( 4 ) zu x t f x kt z e dz

π� ��

� �� (7.128)

EXAMPLE 7.14 (Flow of heat in a semi-infinite medium). Solve the heat conductionproblem described by

2

2

0

PDE: , 0 , 0

BC: (0, ) , 0

IC: ( , 0) 0, 0

u uk x t

tx

u t u t

u x x

� � � � �

� �

� � � �

� ��

u and /u x� � both tend to zero as .x ��



Solution Since u is specified at 0,x � the Fourier sine transform is applicable to thisproblem. Taking the Fourier sine transform of the given PDE and using the notation.

0

2( , ) ( , )sinsU t u x t x dxα α

π�

� �we obtain from Eqs. (7.74) and (7.75) the relation

0

22( , ) [ ( , ); ] ( , )

x

ss

Uk u x t u x t x t

t� � � �

� �

� ��

��

�

or

20

2ss

dUk U k u

dt� �

�

� � (7.129)


202( , ) (1 )k t

su

U t e ��

� �

�

� � (7.130)

Inverting by Fourier inverse sine transform, we obtain

0

2( , ) ( , ) sinsu t U t x dα α α α

π�

� �Therefore,

2

00

2 sin( , ) (1 )k tx

u x t u e dαα απ α

� �� (7.131)

Noting that

2

0

2erf ( )

�

�

� �y uy e du

and using the standard integral

2

0

sin (2 )erf ( )

2

ye d yα α πα

α� � ��

we have solution (7.131) in the form

02( , ) erf

2 2 2

� �

�

� ��

u xu x t

kt(7.132)

Finally, the solution of the heat conduction problem is

2/ 2

0 00

2( , ) 1 erfc

2�

�

� � � ��

x kt u xu x t u e du u

kt(7.133)



EXAMPLE 7.15 Determine the temperature distribution in the semi-infinite medium 0,x �

when the end 0x � is maintained at zero temperature and the initial temperature distribution

is ( ).f x

Solution The given problem is described by

2

2PDE: 0 , 0

u uK x t

t x� � � � ��

� �(7.134)

BC: (0, ) 0, 0u t t= > (7.135)

IC: ( , 0) ( ), 0� � � �u x f x x (7.136)

and, , / ,u u x� � both tend to zero as .x �� Taking the Fourier sine transform of Eq. (7.134)

and denoting

[ ( , ); ] bys su x t x U��we have

2

20 0

2 2sin sin

u ux dx K x dx

t xα α

π π� �

��

which becomes

2[ (0, ) ]ss

dUK u t U

dt� ��

Using the BC (7.135), we obtain

2 0�� s

sdU

K Udt

(7.137)

Also, taking the Fourier sine transform of IC (7.136), we get

( ) at 0s sU F t�� (7.138)

Now, Eq. (7.137) can be rewritten as

2( ) 0K t

sd

U edt

α = (7.139)

Integrating, we get

2const .K t

sU e ��

Using Eq. (7.138), we note that ( ) constant� �sF . Therefore,

2

2

( )

( )

�

�

�

��

�

�

K ts s

K ts s

U e F

U F e (7.140)



Finally, taking the inverse Fourier sine transform of Eq. (7.140), we obtain

2

0

2( , ) ( ) sinK t

su x t F e x dαα α απ

� ��

�� 5�+��4��

Wave motions that occur in nature, viz., sound waves, surface waves, transverse vibrations ofan infinite string, and of mechanical systems are governed by the wave equation. As our firstexample, we shall consider the transverse displacements of an infinite string.

EXAMPLE 7.16 Compute the displacement ( , )u x t of an infinite string using the method

of Fourier transform given that the string is initially at rest and that the initial displacement

is ( ), .f x x��

Solution Displacement of an infinite string is governed by the PDE

2 22

2 2,

u uc x

t x� ��

� �(7.141)

and ICs

( , 0) ( ),u x f x x� �� (7.142)

( , 0) 0tu x � (7.143)

In view of two ICs, the given problem is a properly posed problem. Taking the Fouriertransform of PDE, we have

2 2 2

2 2

1

2 2i x i xu c u

e dx e dxt dx

α απ π

� �

��

�2

2 22

1( , )

2i xue dx c U t

tα α α

π�

��

�i.e.,

22 2

20

d Uc U

dt�� (7.144)


( , ) cos ( ) sin ( )U t A c t B c t� � �� (7.145)

The Fourier transform of the ICs gives

( , ) 0, ( )dU

t U Fdt

� �� (7.146)



i.e., 0sin ( ) cos ( ) 0,tAc c t Bc c t� � � ��

� � � implying 0 or 0.Bc B� � � Also, Eqs. (7.145) and

(7.146) yield

( )A F ��

(7.147)

Thus,

( , ) ( ) cos ( )U t F c t� � �� (7.148)

Taking its inverse Fourier transform, we obtain

1( , ) ( )cos ( )

2i xu x t F c t e d��

��

� � (7.149)

If we simplify Eq. (7.149) further, an interesting result emerges, i.e.,

( )

( )

1( , ) ( )

2 2

1 1( ) ( )

2 2

1 1 1( )

2 2 2

1 1( )

2 2

ic t ic ti u i x

isu icst icst isx

isu is x ct

isu is x ct

e eu x t f u e du e d

f u e du e e e ds

f u e e ds du

f u e e ds

� ��

�

�

� �

� �

��

� � � ��

� ��

��

� ��

� ��

� � � � ��

� ��

� �

� �

� �

� du�

��

��

Using the Fourier integral formula, we arrive at the result

1( , ) [ ( ) ( )]

2u x t f x ct f x ct� � � �

which is the well-known D’Alembert’s solution of the wave equation.

EXAMPLE 7.17 Obtain the solution of free vibrations of a semi-infinite string governed by

2PDE: , 0 , 0tt xxu c u x t� � � � � (7.150)

ICs: ( , 0) ( )�u x f x (7.151)

( , 0) ( )tu x g x� (7.152)

Solution Taking the Fourier sine transform of PDE, we have

2 22

2 20 0sin sin

u ux dx c x dx

t xα α

� ��

� �(7.153)



Now consider

2

20sin

ux dx

xα

�

� ��


2

00sin cos sin

ux u x u x dx

xα α α α α

� ��

Now, since the string is fixed at 0x � for all t and we assume that u and /u x� � both tend

to zero as ,x � � we arrive at

22 2

20 0sin sin

ux dx u x dx U

xα α α α

� ��

�Hence, Eq. (7.153) reduces to

22 2

20

d Uc U

dt�� (7.154)

Its general solution is known to be

( ) cos ( ) ( ) sin ( )U A c t B c t� � � �� (7.155)

where ( ) and ( )A B� � have to be determined. Now, at 0,t � we have

0

0

( )sin ( )

( ) sin ( )

�

�

� �

� �

U f x x dx F

Ug x x dx G

t

α α

α α

�

��

In the solution (7.155), if we take 0,t � then we have

( ) ( )

( ) ( )

U A F

Uc B G

t

α α

α α α

� �

� ��

Substituting ( ) and ( )A B� � into Eq. (7.155), we get

( )( ) cos ( ) sin ( )

��

�

� �G

U F c t c tc



Taking the inverse Fourier sine transform of this relation, we obtain

0

0

0

0

2( , ) ( , ) sin

2 ( )( ) cos ( ) sin sin ( ) sin

1( ) [sin ( ) sin ( )]

2

1 ( )[cos ( ) cos ( )]

2

u x t U t x d

GF c t x c t x d

c

F x ct x ct d

Gx ct x ct d

c

α α απ

αα α α α α απ α

α α α απ

α α α απ α

�

�

�

�

�

� ��

� � � �

� � � �

�

�

�

�Since

0

0

2( ) ( ) sin ( )

2( ) ( ) sin

f x ct F x ct d

g u G u d

α α απ

α α απ

�

�

� � �

�

�

�

0

0

0

2( ) ( ) sin

2 cos( )

2 ( )[cos ( ) cos ( )]

x ct x ct

x ct x ct

x ct

x ct

g u du G d u du

uG d

Gd x ct x ct

α α απ

αα απ α

α α α απ α

� � �

� �

��

�

�

�

� ��

� � � �

� � �

�

�we arrive at the solution

1 1( , ) [ ( ) ( )] ( )

2 2

�

�

� � � � � �x ct

x ctu x t f x ct f x ct g u du

c(7.156)

��% ��4��

One of the most important PDEs that occurs in many applications is the Laplace equation.Steady-state heat conduction, the electric potential in the steady flow of currents in solidconductors, the velocity potential of inviscid, irrotational fluids, the gravitational potential atan exterior point due to ellipsoidal Earth and so on, are all governed by Laplace equation. Weshall now consider a few related examples.



EXAMPLE 7.18 Solve the following boundary value problem in the half-plane y > 0,described by

PDE: 0, , 0

BCs: ( , 0) ( ), ,xx yyu u x y

u x f x x

� � ��

u is bounded as ;y � � u and /u x� � both vanish as | | .x � �

Solution Since x has an infinite range of values, we take the Fourier exponentialtransform of PDE in the variable x to get

[ ; ] [ ; ] 0xx yyu x u x� ��

Since u and /u x� � both vanish as | | ,x �� we have

2 1( , ) 0

2i x

yyU y u e dxαα απ

�

��

22

2

1( , ) ( , ) 0

2i xU y u x y e dx

yαα α

π�

��

� � � ��

i.e.,2

22

( , )( , ) 0

d U yU y

dy

�� (7.157)


( , ) ( ) ( )� ��

�

� �y yU y A e B e (7.158)

Since u must be bounded as , ( , )y U yα�� and its Fourier transform also should be bounded

as ,y � � implying ( ) 0A � � for 0;� � but if 0, ( ) 0;B� �� thus for any ,�

| |( , ) constant ( )��

�

�

yU y e (7.159)

Now the Fourier transform of the BC yields

( , 0) [ ( ); ] ( )U f x x F� � �� (7.160)

From Eqs. (7.159) and (7.160), we find that

( ) const .F � �(7.161)

Hence,

| | | |1( , ) ( ) ( )

2y y i xU y F e f x e e dxα α αα α

π��

� � �



Taking its Fourier inverse transform, we obtain, after replacing the dummy variable x by ,�

the equation

| |1 1( , ) ( )

2 2

1( ) exp [{ [ ( )] | | }]

2

� ��

� � � � � ��

� ��

��

� �

��

� ��

� � �

� �

� �

y i i xu x y f e e d e d

f d i x y d (7.162)

But1

exp { [ ( )] | | }2

i x y dα ξ α απ

�

��

0

0

0

0

2 2

1 1exp { [ ( )]} exp { [ ( )]}

2 2

1 exp{ [ ( )]} 1 exp{ [ ( )]}

2 ( ) 2 ( )

1 1 1

2 ( ) ( )

1

( )

� � � � � ��

� � � ��

� � �

� �

�

��

�

��

� � � � � � �

� � � � ��

� ��

��

� �y i x d y i x d

y i x y i x

y i x y i x

y i x y i x

y

x y (7.163)

Substituting Eq. (7.163) into Eq. (7.162), the required solution is found to be

2 2

( )( , )

( )

y f du x y

x y

ξ ξπ ξ

�

��

� �� (7.164)

This solution is a well-known Poisson integral formula and is valid for 0,y � when ( )f x is

bounded and piecewise continuous for all real x.

EXAMPLE 7.19 Solve the following Neumann problem described by

PDE: ( , ) ( , ) 0, , 0

BC: ( , 0) ( ),

xx yy

y

u x y u x y x y

u x f x x

� � ��

� ��

is bounded asu y ��

and / both vanish as | |u u x x ��



Solution Let us define a function ( , ) ( , ).yx y u x y� � Then

( ) 0xx yy xx yyu uy

φ φ� � � ��

(7.165)

BC: ( , 0) ( , 0) ( )� � �yx u x f x (7.166)

Thus, the function ( , )x y� is a solution of the problem described in Example 7.18 and,

therefore, the solution of Eq. (7.165) subject to Eq. (7.166) is of the form

2 2

( )( , ) , 0

( )

y f dx y y

x y

ξ ξφπ ξ

�

��

� �� (7.167)

However,

2 2

1( , ) ( , ) ( )

( )

y dyu x y x y dy f d

x yφ ξ ξ

π ξ

�

��

� �� Therefore,

2 21( , ) ( ) log [( ) ] const .

2u x y f x y dξ ξ ξ

π�

�� (7.168)

is the required solution.

��' &��6�&��

EXAMPLE 7.20 Find the Fourier transform of the normal density function

2 2 2( ) exp [ ( ) /2 ] / 2� �� f x x m

Solution From the definition of Fourier transform we have

2 2

2

1 exp [ ( ) /2 ]( )

2 2

i xx me dx��

� ��

�

��

� ��

Let ;x m

z�

�

� then

2/2( )( )

2

zi m ze

e dz� ��

��

� ��



Expanding i ze �� in the form

2/2

0

( )

2 !

z ni m

n

e i ze dz

n� ��

�

� ��

�� (7.169)

and denoting

2/2

, 0,1, 2,2

n z

nz e

I dz nπ

��

�� (7.170)

we obtain

0

( )2 ( )

!

ni m

nn

iF e I

nα ασπ α

�

�� (7.171)

For 0,n � we have

2

2

22 2/ 2

2 1 20 1 2

/22

0 0

exp ( /2) exp ( /2)( )

2 2

[in polar coordinates]2

1

z

r

z zeI dz dz dz

er dr d

−

− − −

−

− −= =

=

=

�

� �

��

� � �

� � �

�

� ��

To compute , 0,nI n � we can integrate by parts to get

2 21 /2 2 /222 ( ) ( 1) ( 1) 2n z n z

n nI z d e n z e dz n I� ��

� ��

� ��

Thus, we get a recurrence formula

2( 1)n nI n I�

� �

Finally, we note that 0,nI � when n is odd, since 2/2n zz e−⋅ is an odd function. In that case

we have

2 2 2

2 4

(2 1)

(2 1) (2 3)

(2 1) (2 3) ... 3 1

n n

n

I n I

n n I

n n

−

−

= −

= − −

= − − ×

�



However,

(2 ) (2 1) (2 2) ... 3 2 1(2 1) (2 3) ... 3 1

(2 ) (2 2) ... 4 2

(2 )!

2 !n

n n nn n

n n

n

n

� � � ��

� �

�

Finally, Eq. (7.171) becomes

2 2 22 2

0 0

( ) (2 )! 12 ( ) exp ( / 2)

(2 )! ! 22 !

nni m i m i m

nn n

i nF e e e

n nnα α αασ α σπ α α σ

� �

� �

� ��

Thus,

2 2exp ( /2 )( ) [ ( ); ]

2

i mF f x

� � ��

EXAMPLE 7.21 Using the Fourier cosine transform, find the temperature ( , )u x t in a semi-

infinite rod 0 ,x� � � determined by the PDE

, 0 , 0t xxu ku x t� � � � �

subject to

0

IC: ( , 0) 0, 0 ,

BC: (0, ) (a constant) when 0 and 0x

u x x

u t u x t

� � � �

� � � �

u, /u x� � both tend to zero as .x ��

Solution Since /u x� � is given at the lower limit, we take the Fourier cosine transform

of the given PDE, to obtain

2

20 0

00

2 2cos cos

2 2cos sin

u ux dx k x dx

t x

u uk x k x dx

x x

α απ π

α α απ π

∞ ∞

∞ ∞

=

= +

� ��

� ��

� �

�

From physical considerations, we expect / 0 as .u x x� � �� Therefore,

20

00

2 2 2( ) ( sin ) cosc

x

d uU k k u x k u x dx

dt xα α α α

π π π��

�

��



If 0 as ,u x� � � we have

20

2( )c c

dU ku k U

dt�

�

� �

20

2( )c c

dU k U ku

dt�

�

� �

which is a linear ODE, whose general solution is found to be of the form

2 2

02

( )� �

�

�

k t k tc

de U ku e

dt


2 2

02

const .� �

�

� ��k t k t

ce U ku e dt (7.172)

Taking the cosine transform of the IC, we get

0�cU when 0�t

Using this condition in Eq. (7.172), we have

02

20 const .

u

� �

� �

which yields

02

2const .

u

��

� �

Hence,

202

2(1 )�

� �

�

� �

k tc

uU e (7.173)

Finally, taking the Fourier inverse cosine transform of Eq. (7.173), we have

2020

2 cos( , ) (1 )k tu x

u x t e dαα απ α

� ��

EXAMPLE 7.22 Using the method of integral transform, solve the following potentialproblem in the semi-infinite strip described by

PDE: 0, 0 , 0xx yyu u x y a� � � � � � �



subject to

BCs: ( , 0) ( )

( , ) 0

( , ) 0, 0 , 0

u x f x

u x a

u x y y a x

�

�

� � � � � �

and

/u x� � tends to zero as .x ��

Solution Taking the Fourier sine transform of the above PDE, we obtain

2 2

2 20 0

2 2sin sin 0

u ux dx x dx

x yα α

π π� �

� ��


2

200

2 2sin cos 0sd Uu u

x x dxx x dy

α α απ π

� � � � ��

Using the fact that / 0 asu x x� �� and integrating again by parts, we obtain

22

0 20

2 2( cos ) sin 0sd Uu x u x dx

dyα α α α

π π��

Using the BC

( , ) 0, 0 , 0u x y y a x � ��

we have

22

20��

ss

d UU

dy


( ) cosh ( ) sinh� � � �� sU A y B y (7.174)

Now, the Fourier sine transform of the first two BCs gives

0

2( , 0) ( ) sin , ( , ) 0s sU f d U aα ξ αξ ξ α

π�

� ��Using these two relations in Eq. (7.174), we obtain

0

2( ) ( ) sin

0 ( ) cosh ( ) sinh

A f d

A a B a

α ξ αξ ξπ

α α α α

��

� �

� (7.175)



Therefore,

0

cosh 2( ) ( ) sin

sinh

aB f d

x

αα ξ αξ ξα π

�� (7.176)

Thus, from Eqs. (7.174) to (7.176), we obtain

0 0

0

2 cosh sinhcosh ( ) sin ( ) sin

sinh

2 sinh ( )( ) sin

sin

sa y

U y f d f da

a yf d

a

α αα ξ αξ ξ ξ αξ ξπ α

αξ αξ ξπ α

� �

�

� ��

��

� �

� ��

Finally, taking the inverse Fourier sine transform of Eq. (7.177), we have

0 0

2 sinh ( )( , ) ( ) sin sin

sin

a yu x y f d a x d

a

αξ ξ ξ α απ α

� � ��

EXAMPLE 7.23 Using finite sine transform to solve the following BVP described by

0

0

1PDE: , 0 , 0

BC: (0, ) ( , ) 0 for all

2 , 0 /2IC: ( , 0)

2 1 ,2

� � � �

� �

� � ��

xx tu u x L tk

u t u L t t

xu x L

Lu x

x Lu x L

L

Solution Taking the finite Fourier sine transform of the given PDE, we have

2

20 0

1sin sin

L Lu n x u n xdx dx

L k t Lx

π π��

��(7.178)

Integrating the left-hand side by parts, we get

00sin cos

L Lu n x n u n xdx

x L L x L

π π π� � ��

Again integrating by parts, we obtain

2 2

200

2 2

2

cos sin ( ) { (0, ) ( , ) cos }

( ) [after using the BCs]

L L

s

s

n n x n n x n nu u dx U n u t u L t n

L L L L LL

nU n

L

π π π π π π π

π

⎡ ⎤⎛ ⎞− + = + −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

= −

∫



Dropping the subscript s, Eq. (7.178) becomes

2 2

20

dU knU

dt L

�� (7.179)


2 2

2( , ) exp

��

� �s

knU U n t A t

L(7.180)

where A is a constant to be determined from the ICs. Taking the finite Fourier sine transformof the given IC, we have

/2

0 00 /2

/2 /200

00

00

/2/2

( , 0) 2 sin 2 1 sin

2 12 cos cos

2 12 1 cos cos

/

L L

sL

L L

LL

LL

x n x x n xU n u dx u dx

L L L L

ux n xn x nu dx

L LL L n L L

ux n xn x nu dx

L LL L n L L

π π

ππ ππ

ππ ππ

⎛ ⎞= + −⎜ ⎟⎝ ⎠

⎡ ⎤= − +⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞+ − − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫ ∫

∫

∫or

02 2

4( , 0) sin

2su L n

U nn

�

�

�

From Eq. (7.180) when 0,t � we get

02 2

( , 0) and, therefore,

0 if is even

4 (2 1)sin if is odd

2(2 1)

�

�

�

��

� ��

sA U n

n

u LA rn

r

i.e.

02 2

( 1) 4

(2 1)

r u LA

r �

��

�

Thus, from Eq. (7.180), we have

2 20

2 2 2

( 1) 4 (2 1)( , ) exp

(2 1)

r

su L k r t

U n tr L

�

�

� ��

� � (7.181)



Applying the inversion theorem to Eq. (7.181), we obtain

2 202 2 2

1 0

82 ( 1) (2 1) (2 1)( , ) ( , ) sin exp sin

(2 1)

� � ��

� �

� �

� ��

� � � �

r

sn r

un x k r r xU x t U n t t

L L Lr L

EXAMPLE 7.24 Find the steady-state temperature distribution ( , )u x y in a long square bar

of side � with one face maintained at constant temperature 0u and the other faces at zero

temperature.

Solution Mathematically, the above problem can be posed as the following BVP:

0

PDE: 0, 0 , 0

BCs: (0, ) ( , ) 0

( , 0), ( , )

xx yyu u x y

u y u y

u x u x u

π π

π

π

+ = < < < <

= =

=

Taking the finite Fourier sine transform with respect to the variable x, we have

2 2

2 20 0

2

20 00

22

0 20

sin sin 0

sin cos sin 0

( cos ) sin 0s

u unx dx nx dx

x y

u unx n nx dx u nx dx

x x y

d Un u nx n u nx dx

dy

� ��

� � ��

� �

� � � � ��

� � � �

� �

� �

�

� �

� � �

��

Therefore,

22

20s

sd U

n Udy

� �


cosh sinh� �sU A ny B ny (7.182)

Taking the finite Fourier sine transform of the second set of BCs, we have

0 0 00 0

( , 0) 0

cos 1 cos( , ) sin

��

�

�

��

s

s

U n

nx nU n u nx dx u u

n n



Using these results in Eq. (7.182), we obtain 0A � and

01 cos

sinh�

��

� �n

B n un

Hence,

0 1 cossinh

sinh

�

�

�� s

u nU ny

n n

Finally, taking the finite Fourier sine inverse transform, we have the result

0

1

2 1 cos( , ) sinh sin

sinhn

u nu x y ny nx

n n

ππ π

�

�

�� Thus, the required temperature distribution is

0

0

4 sinh (2 1) sin (2 1)( , ) when odd

(2 1) sinh (2 1)

0 when even

r

u r y r xu x y n

r r

n

π π

�

�

� ��

� �

� �

�

EXAMPLE 7.25 A one-dimensional infinite solid, ,x�� is initially at temperature

F(x). For times 0,t � heat is generated within the solid at a rate of ( , )g x t units. Determine

the temperature in the solid for 0.t �

Solution The problem can be described as follows:

2

2

[ ( , )] 1PDE: , , 0

IC: ( , 0) ( ),

T g x t Tx t

k k tx

T x F x x

� � ��

� ��

� ��

The exponential Fourier transform of PDE with respect to x is given by

2

2

1 1( , )i x i x i xT T

e dx e g x t dx e dxk k tx

α α α� � �

��

��

which on rewriting becomes

2

2

1 1 ( , )( , )i x T dT t

e dx g tk k dtx

α αα�

��

�



or

2 1 1 ( , )( , ) ( , )

dT tT t g t

k k dt

�� (7.183)

The Fourier transform of the IC is given by

( , 0) ( ) ( , 0)i xT e F x dx Fαα α�

��

Therefore,

0( , ) ( , 0) ( ) ( , 0)i xtT t T e F x dx Fαα α α

� ��

� � �� Using this result, Eq. (7.183) can be modified to

2( , )( , ) ( , )

dT tk T t g t

dt

α α α α+ =

The solution of this equation is found to be

22exp ( , ) k tT k dt g t e dt cαα α⎛ ⎞ = +⎜ ⎟⎝ ⎠∫ ∫Using the IC, we get

( , 0)c F ��

Hence, using the IC, the solution of Eq. (7.183) is obtained as

2 2

0( , ) ( , 0) ( , )� ��

��

��

� �� tk t k t

tT t e F g t e dt

Now, taking its inverse exponential Fourier transform, we get

22

0

1( , ) exp ( ) ( ) ( , )

2

t k t

tT x t k t i x F e g t dt dα

αα α α α α

π� ��

� �� (7.184)

where

( ) ( )i x

xF e F x dxαα

� ��

� � ��

( , ) ( , )i x

xg t e g x t dxαα

� ��

�� After changing the order of integration and rewriting, Eq. (7.184) becomes

21( , ) ( ) exp [ ( )]

2 xT x t F x dx k t i x x d

αα α α

π� �

��



2

0

1( , ) exp [ ( ) ( )]

2

t

t xdt g x t dx k t t i x x d

αα α α

π� �

� ��

Now,

2 22 1exp [ ( ) ( )] exp

2 2

x x x xk t i x x d i kt d

kt ktα αα α α α α

� �

��

� �� Let

2� �

��

x xi kt i

kt

Then the above equation can be written as

2 2 21exp [ ( )] exp [ ( ) / (4 )] exp ( )k t i x x d x x kt d

ktαα α α η η

� �

��

But,

2(Standard integral)e dnη π

� ��

�� (7.187)

Using Eqs. (7.186) and (7.187), the required temperature is obtained from Eq. (7.185) in theform

21( , ) ( ) exp [ ( ) / (4 )]

4T x t F x x x kt dx

kt��

��

2

0( , ) exp [ ( ) /4 ( )]

4 ( )

t

t

dtg x t x x t t dx

k t t�

��

��

� � � � � ��

� � � ��

EXAMPLE 7.26 A uniform string of length L is stretched tightly between two fixed points

at 0x � and .x L� If it is displaced a small distance � at a point , 0 ,x b b L� � � and

released from rest at time 0,t � find an expression for the displacement at subsequent times.

Solution Let ( , )u x t denote the displacement of the string. Then at 0,t � the equation

for OA is: ( / ) ,��y b x (see Fig. 7.1) while the equation for AB is given by

( )y x Lb L

��

�

Now, the IBVP is described by the PDE:

2tt xxu c u�



BCs: (0, ) ( , ) 0, 0,

, 0ICs: ( , 0)

( ),

( , 0) 0t

u t u L t t

xx b

bu xx L

b x Lb L

u x

ε

ε

= = ≥

⎧ < <⎪⎪= ⎨ −⎪ < <⎪ −⎩=

Taking the finite Fourier sine transform of the PDE, we have2 2

22 20 0

sin sinL Lu n x u n x

dx c dxL Lt x

π π��

� �or

22

2 00

2

00

2 2 2 2

2

2 2 2

2

sin cos

cos sin

{ (0, ) ( , ) cos }

(after using BCs)

L Ls

L L

s

s

d U u n x n u n xc dx

x L L x Ldt

n n x n n xc u u dx

L L L L

n c n cU u t u L t n

LL

n cU

L

π π π

π π π π

π π π

π

� ��

� ��

��

��

�

�

� ��

Therefore,

2 2 2 2

2 20

��

ss

d U n cU

dt LIts general solution is found to be

( , ) cos sin� �

� �sn ct n ct

U n t A BL L

(7.188)

Now, taking finite Fourier sine transform of ICs, we obtain

0

00

0

2 2

2 2 2 2

( )( , 0) sin sin

cos cos

( ) cos cos( ) ( )

sin sin( )

� � � �

� � � �

� �

� � � �

� �

� � � �

� �

��

�

� ��

� ��

� ��

� �

�

�

b L

sb

b b

L L

b

x n x x L n xU n dx dx

b L b L L

L n x L n xx dx

b n L b n L

L n x L n xx L dx

n b L L n b L L

L n b L n b

L Ln b n b L



or

3

2 2( , 0) sin

( )s

L n bU n

Ln b L b

ε ππ

=−

From Eq. (7.188), when 0,t � we have

3

2 2( , 0) sin

( )s

L n bU n A

Ln b L b

ε ππ

= =−

Further, taking finite Fourier sine transform of the second IC, we get

0�

sdU

dt

From Eq. (7.188) it can be easily seen that 0.B � Thus, we obtain from Eq. (7.188) therelation

3

2 2( , ) sin cos

( )s

L n b n ctU n t

L Ln b L b

ε π ππ

=−

Finally, inverting, we have the required, displacement as

2

2 21

2 1( , ) sin sin cos

( ) n

L n b n x n ctu x t

L L Lb L b n

ε π π ππ

�

��

� �

�6��

1. Find the Fourier transform of

21 , | | 1( )

0, | | 1

� � ��

��

x xf x

x

and hence evaluate

30

cos sincos

2

x x x xdx

x

� ��

2. Find Fourier sine and cosine transforms of .xe�

3. Find the Fourier cosine transform of

2

1( )

1f x

x�

�



4. Using Parseval’s relation for the Fourier cosine transform of

( )

1, 0( )

0,

axg x e

xf x

x

�

�

��

� ��

��

show that

2 2 20

sin 1

2( )

ad e

a a

λλα α πα α

�� 5. Verify the following relations:

(i)0 0

( ) ( ) ( ) ( )s sF G d f x g x dxα α α� �

�� (ii) 2 2

0 0[ ( )] [ ( )]cF d f x dxα α

� ��

(iii) 2 2

0 0[ ( )] [ ( )] .sF d f x dxα α

� ��

6. If 0,a � b is any real or complex, show that

2 2/2 aax bx be dx ea

π� � ��

��7. Solve the problem described by

| |

PDE: ( ) ( ),

BC: Lt ( , ) 0

IC: ( , 0) ( )

t xx

x

u u x t x

u x t

u x x

δ δ

δ��

� � � � � � ��

�

8. If ( , )v x t denotes the solution of

0, , 0

( , 0) ( ),

t xxv v x t

v x f x x

� � ��

� ��

show that

0( , ) ( , )

tu x t v x t dτ τ= −∫



is a solution of

( ), , 0

( , 0) 0,

t xxu u f x x t

u x x

� � ��

� ��

and hence, write down the Green’s function for the above non-homogeneous PDE withthe homogeneous initial condition.

[Duhamel’s principle]

9. The temperature ( , )x t� in the semi-infinite rod 0x � is determined from the differential

equation

22

2t x

θ θα��

subject to

IC: ( , 0) 0

BC: (0, ) ( ) for 0

�

� �

�

� �

x

t t t

Using the Fourier sine transform, derive the solution in the form

22

2 2/(2 )

2( , )

4u

x a t

xx t t e du

uθ φ

π α

� ��

10. If 2 0, for 0 and if ( ) on 0,� � � � �u x u f y x show that by using Fourier transform

technique,

2 2

( )( , )

( )

x f du x y

x y

ξ ξπ ξ

�

��

� ��11. Using the Fourier transform method, show that the solution of the two-dimensional

Laplace equation

2 2

2 20

u u

x y ��

� �

is valid in the half-plane 0,y � subject to the condition

0, 0( , 0)

1, 0

xu x

x

��

��

and

2 2Lt ( , ) 0

x yu x y

� �� in the above half-plane.



12. Using the finite Fourier transform, solve the BVP described by

2

2PDE : , 0 6, 0

V Vx t

t x� � � �

� ��

subject to

BC: (0, ) 0 (6, )

IC: ( , 0) 2

� �

�

x xV t V t

V x x

13. Using the finite Fourier transform, solve the two-dimensional Laplace equation

2 2

02 20 0 , 0

V Vx y y

x yπ� � � � � ��

� �

subject to

0

(0, ) 0, ( , ) 1

( , 0) 0, ( , ) 1.y

V y V y

V x V x y

π= =

= =


BELL, W.W., Special Functions for Scientists and Engineers, Van Nostrand, London, 1968.

CARSLAW, H. and JAEGER, J., Conduction of Heat in Solids, Oxford University Press,Fairlawn, N.J., 1950.

CHESTER, C.R., Techniques in Partial Differential Equations, McGraw-Hill, New York,1971.

CHURCHILL, R., Fourier Series and Boundary Value Problems, 2nd ed., McGraw-Hill,New York, 1963.

COLE, J.D., On a quasi-linear parabolic equation occurring in aerodynamics, Q. Appl.Math., 9, pp. 225–236, 1951.

COPSON, E.T., Partial Differential Equations, S. Chand & Co., New Delhi, 1976.

DENNEMEYER, R., Introduction to Partial Differential Equations, McGraw-Hill, New York,1968.

DUCHALEAU PAUL and ZACHMANN, D.W., Partial Differential Equations (Schaum’sOutline Series in Mathematics), McGraw-Hill, New York, 1986.

DUFF, G.F.D. and NAYLOR, D., Differential Equations of Applied Mathematics, Wiley,New York, 1966.

GARABEDIAN, P., Partial Differential Equations, Wiley, New York, 1964.

GELFAND, I. and SHILOV, G., Generalized Functions, Vol. I, Academic Press, New York,1964.

GREENBERG, M., Applications of Green’s Functions in Science and Engineering,Prentice Hall, Englewood Cliffs, N.J., 1971.

GUSTAFSON, K.E., Introduction to Partial Differential Equations and Hilbert SpaceMethods, Wiley, New York, 1987.

447

Bibliography


448 BIBLIOGRAPHY

HILDEBRAND, F.B., Advanced Calculus for Applications, Prentice Hall, Englewood Cliffs,N.J., 1963.

HOPE, E., The PDE: ut + uux = huxx, Comm. Pure Appl. Math., 3, pp. 201–230, 1950.

JOHN, F., Partial Differential Equations, Springer-Verlag, New York, 1971.

MACKIE, A.G., Boundary Value Problems, Oliver and Boyd, London, 1965.

PRASAD, PHOOLAN and RENUKA, R., Partial Differential Equations, Wiley Eastern,New Delhi, 1987.

SANKARA RAO, K., Classical Mechanics, Prentice-Hall of India, New Delhi, 2005.

SNEDDON, I.N., Elements of Partial Differential Equations, International edition, McGraw-Hill, Singapore, 1986.

SNEDDON, I.N., The Use of Integral Transforms, Tata McGraw-Hill, New Delhi, 1974.

STACKGOLD, IVAR, Boundary Value Problems of Mathematical Physics, Vol. II,Macmillan, New York, 1968.

TYCHONOV, A.N. and SAMARSKI, A.A., Partial Differential Equations of MathematicalPhysics, Vol. 1, Holden-Day, London, 1964.

WEINBERGER, H.F., A First Course in Partial Differential Equations, Plaisdell Publishing Co.,New York, 1965.


��

1. px qy x y� � �

2. 4pq xyz�

3. , 0xy x y

Fz z

��

4. (i) 3 3 2 2( , ) 0F x y x z− − =

(ii) 2 2( 2 2 , ( )) 0F x y x y z x y− + − − =

5. 2 2 4 2 0x y z z� � �

6. 2 2 22 4x y x z z� � � �

7. 2 4 2 2 4( ) ( ) 2 ( ) 2 ( ) 0x y z z x y z z x y z z x y z� � � � � � � � � � � �

8. (2 1), ( 1)t tx s e y s e� � � �

2 ( 1), ( 1)t tp s e q s e� � � �

2 2 25( 1) 3 ( 1)

2t tz s e s e� � � �

and the equation of the integral surface is

(4 3 ) / 2.z y x y� �

449

Answers and Keys to Exercises


450 ANSWERS AND KEYS TO EXERCISES

9. ( ), ( )/22

t t t tsx e e y e e− −= + = −

( )/2, ( )/2t t t tp e e q s e e− −= + = −

2 2

4 4 2t ts s s

z e e��

The equation of the integral surface is

2 2 2(1 )z x y� �

10. 2 (2 ), 2 2 ( 1)t tx s e y s e� �

� � � �

2 2tz s e��

and the equation of the integral surface is24 ( 2 ) 0z x y� � �

11. 212z xy c� �

12. 2 2 2( )z x y c� � �

13. 2 2 2 2( )z a x ay b� � �

14. (i)23/2 3/

2

2 1 1( )

3 9 3xz y a be

x= + + + +

(ii)2

2 34

ax b az

yy y= + −

15. ( /( 1))z ax a a y c� � � �

16. (i) 2 2( 1) ( / )z a x y a b� � � �

(ii) 2 2 3/ 2 2 2 1/ 21( ) ( )

3z x a y a b� � � � �

17. sin ( )z ax by ab� � �

18. (i) The given PDE is of Clairaut’s form

2 2

p qz xp yq

q p

� ��

� �Its complete integral is

2 2

a bz ax by

b a

� ��


ANSWERS AND KEYS TO EXERCISES 451

(ii)4 4a b

z ax byab

� ��

� �

19. The required surface is2 2 3 22 2x y z z� � � � �

20. Similar to Example 0.16.21. The auxiliary equations for the given PDE are

( )

dx dy dz

x y x y z z� �

� � � �(1)

from which we observe that 0.dx dy dz� � �


1x y z c� � � (2)

We also observe that

dx dy dz dz

x y x y z z z

� ��

� � � � � which on integration yields

21

ln ( ) ln ( ) ln2

x y z z c� � � �

or

22( )/x y z z c− + = (3)

Given 1,z � and 2 2 1.x y+ = Therefore, Eqs. (2) and (3) become

1 21 , 1x y c x y c� � � � � �

whose solution is found to be

1 2 1 21 and2 2

c c c cx y

� �� (4)

But, we are given 2 2 1x y� �

Therefore,

2 21 2 1 21 1

2 2

c c c c� ��

� � which simplifies to

2 21 2 1 2 0c c c c� � � �



or

22

2 2( ) ( ) 0

x y z x y zx y z x y z

z z

� � � ��

� �

is the required integral surface.

22. Given / ,u w y= which implies 2/ ,

yx x y

wwu w y u

yy� � � �

Substituting in the given PDE, we get

2y

x

wx ww u y

y yy

⎛ ⎞= + +⎜ ⎟⎜ ⎟⎝ ⎠

That is,

or 0x y x yx w w

w w xw ywy y y

� � � � �

This is a Lagrange’s equation, whose auxiliary equations are

0

dx dy dw

x y� �

�

The first two members give us 0,x dy y dx� � which on integration yields 1.xy c� The

last equation is 0,dw � which on integration gives 2.w c� Hence, the solution

is ( )w f xy� . Therefore, the correct choice is (D).

23. The given differential equation can be recast as

2 2

p qz xp yq

q p

� ��

� �

which is of Clairauts form. Hence, its complete integral is

2 2( )z ax by ab ba� �

� � � �

Therefore, the correct choice is (A)

24. 1 ( ), 1 ( )z z

p yA xy q xA xyx y

′ ′= = + = = +� ��

Eliminating A� in the above pair of equations, we get

px qy x y� � �

Hence, the correct choice is (B).25. The given PDE is in Clairaut’s form. Hence, the correct choice is (B).



��1. The given PDE is hyperbolic if the determinant

2 2 24 4( ) [2 ( ) ] 0B AC x y x y� � � � � �

That is, when 22 ( ) 0x y� � � as 0.x y� � This means that 2( ) 2,x y� � implying

| | 2;y x� � in other words, 2 2.y x� � � � Hence, the given PDE is hyperbolic

between the straight lines 2y x� � and 2,y x� � but not on the line y = x.

2. The discriminant 2 4B AC� is 24(1 ).x� Thus the given PDE is

Hyperbolic Elliptic

If 21 0 or | | 1x x� � � If 21 0 or | | 1x x� � �

The characteristic equations are The characteristic equations are

2

2

4 1

2 1

dy B B AC

dx A x

� ��

�2

1

1

dyi

dx x� � �

�

On integration, we obtain On integration, we get

1siny x c�� 1coshy i x c��

Hence, the characteristic equations Hence, the characteristic equationsare are

1siny xξ �� 1coshy i xξ ��

1siny xη �� 1coshy i xη ��

3. The discriminant 2 4 4 0B AC xy� � � � is true when 0, 0x y� � and when x < 0, y < 0.In either of these cases, the given PDE is of elliptic type and

dyi xy

dx� �

On integration, we get 3/212 (2/3)y ix c� � and 3/2

22 (2/3) .y ix c� � If we put3/2(2/3) ,x�η 1/22 ,y�ξ we get the required canonical form

1 1

3u u u u� � �ξξ ηη ξ ηξ η



4. (a) Hyperbolic in the first quadrant. The characteristic equations are:

2 2 ,x y� �ξ 2 2x y� �η

The canonical form is2 22( ) 0u u uξη ξ ηξ η η ξ− − + =

(b) Parabolic type.

Characteristic equation: ,y x y� ��

Canonical form: 0.u��

(c) Elliptic for finite values of x and y. The canonical form is

uuu u u

��

� ��

(d) Parabolic everywhere. The characteristic equations are:

/ ,y x yξ η= =

Canonical form: 0uηη =

(e) The equation is hyperbolic. The characteristic equations are:

,y xξ � � ( /4)y xη � �

Canonical form:1 8

3 9u uξη η� �

5. The equation is hyperbolic. The characteristic equations are:

13 ,

3y x y x� ��

Canonical equation: 0.u��

The general solution is

( , ) ( 3 )3

xu x y f y x g y

� ��

7.2 2 2

*2 2

( ) ( ) ( ) ( ) ( ) ( ) ( )L v Av Bv Cv Dv Ev Fvx y x yx y

� � � � � ��

9. Comparing with the general second order PDE, that is,

xx xy yy x yAZ BZ CZ DZ EZ FZ G� � � � � �



From the given PDE, we find that 21, 2, cos .A B C x= = = Then, the discriminant is

given by 2 2 24 4 4cos 4sin .B AC x x− = − = Therefore, the given PDE is of hyperbolic

type if sin 0.x � On integration, the required characteristics are found to be

1 2cos and cosy x x c y x x c� � � � � �

10. Let us introduce the transformation x yη = + and .x y� � � Then, ,x x xu u uη ξη ξ= +

y y yu u uη ξη ξ= + and 2 , 2xx yyu u u u u u u u�� , .xyu u u��

Substituting these results in the given PDE, we get 0,u�� which is the requiredcanonical form.

11. Comparing the given PDE with the standard PDE, obtain 3 2, 0, ( 1).A y B C x� � � � �

Therefore, the discriminant of the given PDE is given as 2 3 24 4 ( 1).B AC y x� � �

Evidently, the given PDE is hyperbolic in {( , ), 0}.x y y > Thus, the correct choice is(B).

12. Comparing the given PDE with the standard PDE, we find the discriminant as

2 4B AC− 2 2 2 2 2( 1) 4 ( 1) ( 1)x y x y y y= − − − −

2 2 2( 1) [ 1 4 ( 1)]x y y y y= − − − −

2 2 2( 1)[ 3 4 1] 0.x y y y= − − + − >

Hence, the given PDE is hyperbolic everywhere except along 0,x � that is, exceptalong y-axis. Therefore, the correct choice is (B).

13. Comparing the given PDE with the standard PDE of second order, we find that thecharacteristic equations are given by

2 4,

2

dy B B AC

dx A

� � ��

where 2 2, 0, .A x B C y� � � �

Thus,

2 2

2

4.

2

x ydy y

dx xx� � � �

Taking –ve sign we have,

.dy dx

y x� �


ln ln ln .y x C� � �



That is,,xy c�

which is a rectangular hyperbola. Hence, the correct choice is (A). If we take +vesign, the correct choice is (D).

14. Comparing the given PDE with the standard PDE of second order, we observe that

, 2 ,A y B xy C x� � � . Hence its discriminant 2 2 24 4 4B AC x y xy� � � should be

positive. For hyperbolic case, we should have 1.xy � Hence, the correct choice is(C).

15. (i) uCF = ex�1(y + x) + e2x�2(y + x)(ii) uCF = ex�1(x – y) + e3x�2(2x – y).

16. uCF = 1

i ia x b yi

i

c e�

�

��

where F(ai, bi) = ai2 + aibi + ai + bi + 1 = 0.

17. (i) u = ex�1(y) + e–x�2(y + x) + 1

2sin (x + 2y)

(ii) u = �1(x) + e3x�2(y + 2x) – 1

6ex+2y

(iii) u = ey[y�2(x – 2y) + �1(x – 2y) + ey[y2�3(2x + y) + y�2(2x + y) + �1(2x + y)].

18. (i) u = �1(xy) + x2�2(xy) + y

x

(ii) u = �1(y) + e3x �2(y + 2x) – 1

6ex+2y.

19. (i) u = �1(y – x) + �2(y – 2x) + 2 3

2 3

x y x�

(ii) u = �1(y + ix) + �2(y – ix) – 2 2

1

( )p q� cos px cos qy

(iii) u = �1(y – x) + �2(y + 2x) + yex

(iv) u = �1(2y + x) + x�2(2y + x) + 2x2 log(x + 2y)

(v) u = �1(y + x) + �2(y + 2x) + 1

4e2x+3y –

1

15sin(x – 2y).

��

1. It has been shown in Example 2.4 that the possible solution of the given BVPsatisfying all the BCs except the first one is

1

( , ) sinnn

n

u r B r n� �

�

��



Using the BC:

2400(10, ) ( )u � ��

��

we have

10 sin sinnn nB n b nθ θ=∑ ∑

where

2

0

2 400( ) sinnb n d

ππθ θ θ θ

π π= −∫

2

2 0

800( )sin for oddn d n

ππθ θ θ θ

π= −∫


2 3 3 2 3

800 2 2 160010 ( 1) [1 ( 1) ] for oddn n n

nB nn n nπ π

⎡ ⎤= − − = − −⎢ ⎥⎣ ⎦

Therefore,2 3 2 1(1600 2)/ (2 1) 10 n

nB nπ −= × −

Hence the required solution is

2 1

2 31

3200 1( , ) sin (2 1)

10(2 1)

n

n

ru r n

n� �

�

�

�

� ��

��

2. The problem reduces to the solution of

2PDE: 0T

rr r⎛ ⎞ =⎜ ⎟⎝ ⎠

� ��

subject to

1 2BCs: ( ) , ( ) 0 atT

T a T h T T r br

= + − = =��

Integration of PDE with respect to r gives

12

cT

r r=�

�(i)

Using the second BC, we obtain

12 2

( )bc

h T Tb

− = (ii)



Integrating again, we get from equation (i) the relation

12

cT c

r� � � (iii)

Using the first BC, we have

11 2

cT c

a� � � (iv)

From equations (ii)–(iv),

1 1 12 2 12

( )c c c

hT hT b hT h Tb ab

� ��

which gives

2 11

[ ]

( )

h T T abc

a hb b a

��

�

� �

Finally, the required steady temperature T can be obtained from equations (iii) and(iv) as

2 21 2 2 1

1( ) ( )

[ ( )] [ ( )]

h T T ab h T T abT T

r a hb b a a a hb b a

� ��

� � � �

22 1

1( ) 1 1

[ ( )]

h T T abT

a hb b a a r

� � ��

3. It has been shown in Example 2.15 that

1

0

( , ) ( / ) (cos )n nn n n

n

T r A r B r P� ��

�

� ��

using the BCs:

11

0

( / ) (cos )n nn n n

n

T A a B a P ��

�

� ��

For 0,n �

1 0 0/T A B a= + (i)

For 1,n �

21 10 /A a B a= + (ii)



Also using the second BC, we have

12(1 cos ) ( / ) (cos )n n

n n nT A b B b Pθ θ�� For n = 0,

2 0 0 /T A B b� �

For n = 1,

22 1 1/T A b B b� � �

Solving (i) and (iii), we get

1 2 1 20 2 02

,1/ 1/

T T T TA T B

a bb ab

− −= − =−−

Similarly, solving (ii) and (iv), we obtain 1 1andB A . For 2,3,...,n � we get the followinghomogeneous system

1/ 0,n nn nA a B a ++ =

1/ 0n nn nA b B b ++ =

which gives 0n nA B� � for 2,3,...n �4. Solve the PDE:

2 2

2 2 2

1 1T T T

r rr r θ+ +� � �

�� subject to the BCs:

( , ) 0 , ( , ) 100 , 0 2T a T b� � � ��

The possible solution is

1 2( ) lnT r c r c� � (i)

Using the BCs, we obtain 1 20 ln .c a c� � Therefore, 2 1 ln .c c a= − Also,

1 1 1100 ln ln ln ( / )c b c a c b a= − =

which gives

1 2100 100ln

,ln ( / ) ln ( / )

ac c

b a b a

−= =

Hence the temperature distribution in the annulus is obtained from equation (i) as

ln ( / )100

ln ( / )

r aT

b a=



5. Following the interior Dirichlet’s problem, the solution of Laplace equation in polarcoordinates is

0( , ) ( cos sin )2

nn n

Au r r A n B nθ θ θ= + +∑ (i)

Using the prescribed BC, we have

0( , ) ( cos sin )2

nn n

Au a a A n B nθ θ θ= + +∑ (ii)

which is a full-range Fourier series, where

00

1 cA c d

α αθπ π

= =∫

0

1 sincosn

nc n

a A c n dn

α αθ θπ π

= =∫Similarly,

0

1 cos 1sinn

nc n

a B c n dn n

α αθ θπ π

⎛ ⎞= = − −⎜ ⎟⎝ ⎠∫Thus the temperature distribution at the interior points of the disc can be obtainedfrom equation (i) as

sin cos sin

( , ) sin cos2 1

n

n

c r c c n c n nu r n n

n n na

α θ α θθ α θπ π π π

⎡ ⎤= + + −⎢ ⎥⎣ ⎦∑

[sin ( ) sin )2

nc r c

n na n

α α θ θπ π

⎛ ⎞= + − +⎜ ⎟⎝ ⎠∑or

1

sin ( ) sin( , )

2

n

n

c r n nu r

a n

α α θ θθπ

�

�

� ��

6. Solve the PDE:

2 0ψ∇ = (i)

subject to the BCs:

, 0� � �� (ii)

, nnrθ β ψ α= =∑ (iii)



Using spherical polar coordinates and in view of symmetry of the cone, the solutionof equation (i) is

( 1)1 2 3 4[ ] [ (cos ) (cos )]n n

n nc r c r c P c Q� � �� (iv)

When 0,r �� must be finite, implying 2 0.c � Hence the solution may be of the

form

[ (cos ) (cos )]nn nr AP BQ� � ��

Using the BC (ii), we get

(cos )

(cos )n

n

APB

Q

�

�

� �

Therefore,

0

( , ) [ (cos ) (cos ) (cos ) (cos )](cos )

nn

n n n nnn

r Ar P Q P Q

Q� � � � � �

��

� ��

Now, applying the BC (iii), we obtain

[ (cos ) (cos ) (cos ) (cos )]n

nn n n n

AP Q P Q

�

� � � ��

�


0

(cos ) (cos ) (cos ) (cos )

(cos ) (cos ) (cos ) (cos )nn n n n

nn n n nn

P Q P Qr

P Q P Q

� � � ��

� � � ��

� ��

��

7. Let

2 2 2| | ( ) ( ) ( )

q q

x x y y z z� � �

�� r r

then

3

( )

| |

q x x

x

ψ ′−= −′−r r

��

2 2

2 3 5

3 ( )

| | | |

q q x x

x

ψ ′−= − +′ ′− −r r r r

��

Similarly,

2 2

2 3 5

3 ( )

| | | |

q q y y

y

ψ ′−= − +′ ′− −r r r r

��



2 2

2 3 5

3 ( )

| | | |

q q z z

z

ψ ′−= − +′ ′− −r r r r

��

Therefore,

2 0ψ∇ =

Hence,

| |

q� �

��r r

is one of the elementary solutions of the Laplace equation.

8. Assume ( ) ( ),R r F� �� by variables separable method, we have

22r R rR F

R Fλ

′′ ′ ′′+ = − =

Since the velocity components must be periodic with respect to ,� we have

cos sinF c D�� (i)

The ODE for R becomes

2 2 0,r R rR n R n�� (ii)

This is Euler’s homogeneous equation whose solution is

n nR Ar Br−= +

For special case when 0,n� � � the solution of (i) and (ii) are

1 2ln ,R k r k F M N��

Therefore, the general solution of the given PDE is

1 21 1

( ln ) ( ) ( cos sin ) ( cos sin )n nn n n n

n n

k r k M N r A n B n r c n D nφ θ θ θ θ θ−

= == + + + + + +∑ ∑

� �(iii)

Now the BCs

1 11

1

( ) ( cos sin ) ( cos sin )n nr n n n n

n

kv M N nr A n B n nr c n D n

r r

φ θ θ θ θ θ− − −

== = + + + − +∑ ∑

��

To satisfy the BC at infinity, i.e.

cosr

Ur

��

�

� ��

��



we must choose

1 , 0 ( 2), 0n nA U A n B��

Similarly, the BC: 0r a

r

�

�

�

��

gives, for all values of ,� the relations

11

1

0 ( ) cos ( cos sin )nn n

n

kM N U na c n D n

a� � � �

� �

�

� � � � ��

�

implying thereby

21 10, , 0 ( 2), 0n nk c a U c n D��


2

21 cos

aU r k

rφ θ θ�

� ��

9. 2( , )v x y x x� � satisfies 2 2,v� � � and the boundary condition 0v � along 0x � and

1.x � Let .v� �� Now construct a function � such that 2 0,ω∇ = satisfying

2

(0, ) (1, ) 0, 0 1

( ,0) ( ,1) , 0 1

y y y

x x x x x

ω ω

ω ω

= = ≤ ≤

= = − ≤ ≤

Following Example 2.21, it can be shown that

23 3

1

sin[2 1) ] sinh [(2 1) ] sinh [(2 1) (1 )]( , )

sinh (2 1)(2 1)n

n x n y n yx y x x

nn

� � ��

��

� ��

��

10. Since u is a function of andr � alone, in cylindrical polar coordinates, the solution

of Laplace equation satisfying Lt ( , ) 0r

u r ��

�

�is given by


n

u r r A n B n� � ��

� ��

The given boundary condition

1 0( cos sin ) sin 3r a

nn n

uuna A n B n

r aθ θ θ

=

− −= − + =∑��

will be satisfied only if 0nA � for all n. Thus, we have

10 sin 3 sinnn

una B n

aθ θ− −= −∑



This implies that all the constants nB will be zero, except, 3B which is given by

4 033 sin 3 sin 3

ua B

a� �

�

� �

i.e.,

303 3

uB a� �


30( , ) sin 3 ( )3

u au r a r

r� �

� ��

�

12. (1 ) ( )y x xu e Ae Be� � ��

Chapter 3

1. Mathematically, the problem is described as

2PDE: t xxT Tα=

BCs: 0 at 0, for 0xT x x l t= = = ≥

2IC: ( ,0) , 0T x lx x x l= − ≤ ≤

Using the variables separable method, the physically meaningful non-trivial generalsolution is

2 2( , ) ( cos sin ) tT x t A x B x e � �

� ��

� � (i)

2 2( , )

( sin cos ) tT x tA x B x e

xα λλ λ λ λ −= − +�

�(ii)

The first BC gives from (ii), 0.B � The second BC gives from (ii)2 2

( sin ) 0,tA l e � ��

�

� implying sin 0 or / , 1,2,l n l n� � ��


2 2 2 2

1

( , ) cos exp ( / )nn

nT x t A x n t l

l

��

�

� ��

�(iii)

Finally, use the IC:

2 cosnn

lx x A xl

π⎛ ⎞− = ⎜ ⎟⎝ ⎠∑



where

2

0

2( ) cos

l

nn

A lx x x dxl l

π⎛ ⎞= − ⎜ ⎟⎝ ⎠∫2. Solve

2PDE: , 0t xxT T x aα= ≤ ≤

BCs: (0, ) 0, ( , ) 0T t T a t= =

IC: ( , 0) ( )T x f x�

The general solution is

2 2( , ) ( cos sin ) tT x t A x B x e α λλ λ −= + (i)

The first BC gives 0;A � the second BC gives

sin 0, implying , 1, 2,n

a na

πλ λ= = = …

Therefore,

2 2( , ) sin t

nn

T x t B x ea

α λπ −⎛ ⎞= ⎜ ⎟⎝ ⎠∑ (ii)

Now the IC:

( ) sinnn

f x B xa

π⎛ ⎞= ⎜ ⎟⎝ ⎠∑Hence the required solution is

22

1

( , ) sin expnn

n nT x t B x t

a a

� ��

�

� ��

��

where

0

2( ) sin

a

nn

B f x x dxa a

�� 4. The general solution is

2( , ) ( cos sin ) tT x t A x B x e �

� ��

� �

Using the BCs and the superposition principle, the solution is

2 2( , ) sin ( )n t

nT x t B e n xπ π−=∑



Now use the IC:

1

0

1/2 1

0 1/2

1/2 1

0 1/2

2 ( , 0) sin ( )

2 2 sin ( ) 2(1 ) sin ( )

4 4[ cos ( )] (1 ) [ cos( )]

n

n

T x n x dx

x n x dx x n x dx

xd n x x d n xn π

π

π π

π ππ

=

⎡ ⎤= + −⎢ ⎥⎣ ⎦

= − + − −

∫

∫ ∫

∫ ∫

B

Integration by parts gives

1/2 1

2 20 1/2

4 sin( ) sin ( ) 8sin

2nn x n x n

Bn n n n

π π ππ π π π

⎡ ⎤⎧ ⎫ ⎧ ⎫= − − + =⎢ ⎥⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦


2 2

2 2

8 1( , ) sin sin ( )

2n tn

T x t n x en

ππ ππ

−= ∑5. Let ( , ) ( ) ( ).r t R r T t� � Substituting into the given PDE, we get

2Tp

Tν

′= − (i)

2 210, const .R R p R p

r′′ ′+ + = − = (ii)

Integration of (i) gives

21 exp ( )T c p tν= − (iii)

Equation (ii) is a Bessel’s equation of order zero whose solution is

2 0 3 0( ) ( )R c J pr c Y pr= +

Thus the general solution is

20 0( , ) [ ( ) ( )] exp ( )r t AJ pr BY pr p tθ ν= + − (iv)

In view of the first BC, the solution is

20( , ) ( ) exp ( )r t AJ pr p tθ ν= −

The second BC gives 0 ( ) 0J pa � which has an infinite number of zeros,

say, 1 2, , , ,nλ λ λ… … Therefore,

, 1, 2,np na

λ= = …



Using the principle of superposition, we get the solution

2

0 2( , ) expn n

nr t A J r ta a

� ��

� ��

Now using the IC, we get

2 20( )

4n

nP

a r A J ra

λμ

⎛ ⎞− = ⎜ ⎟⎝ ⎠∑ (v)

which is a Fourier-Bessel series. Multiplying both sides by 0 ( / )mrJ r aλ and integrationfrom 0 to a and using the orthogonality property.

20 0 201

0 if

( ) ( )( ) if

2

a

m nm

n m

rJ r J r dr aJ a n m

λ λλ

≠⎧⎪= ⎨

=⎪⎩∫

we obtain

2 202 2 0

1

( )2 ( )

a mm

m

rPA r a r J dr

aa J

λμ λ

⎛ ⎞= − ⎜ ⎟⎝ ⎠∫which on evalution reduces to

2

31

2

( )m

m m

PaA

J� � ��


22 2

031 1

2( , ) exp ( / )

( )m

mm m m

Par t J r t a

aJ

��

� � ��

� ��

��

6. In view of spherical symmetry, the governing PDE:

2 2t rr rT c T T

r� ��

(i)

Setting ,v rT� we have

2, r

t t rr

v r vvv rT T

r r

��

2

4

{( ) )} 2 ( )r r r rrr

v r v r r v r vT

r

� � �

�

Equation (i) reduces to

2t rrv c v� (ii)



The corresponding boundary and initial conditions are

(0, ) ( , ) 0v t v R t= = (iii)

( ,0) ( )v r rf r� (iv)

respectively. The possible solution of (ii) using BCs is

2 2 2 2( , ) sin exp ( / )nn

v r t B r c n t RR

��

� �� (v)

Now using the IC, we get

1

( ) sinnn

nrf r B r

R

�

�

� ��

�

which is a half-range Fourier series of ( ).rf r Hence

0

2( ) sin

R

nn r

B rf r drR R

π= ∫ (vi)

The required solution is

2 2 2

21

( , ) 1( , ) sin expn

n

r t n c n tT r t B r

r r R R

ν π π�

�

� �� (vii)

8. The governing differential equation is

1, 0 , 0t rr rT k T T r a t

r� ��

BC: 0 (thermally insulated)r a

T

r=

=��

IC: ( ,0) ( )T r f r=

The required solution is

20

1

( , ) ( ) exp ( )n n nn

T r t A J r k t� �

�

� ��

where

00

20

0

( ) ( )

( )

a

n

n a

n

rf r J r drA

rJ r dr

λ

λ=∫∫



9. The formal solution is found to be

3

( , ) ( , ) ( )R

u t k t f dα′ ′ ′= −∫∫∫r r r r r

where

23/2 | |

( , ) (4 ) exp4

k t tt

παα

− ⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠

rr

Here, the function ( , )k tr is called a fundamental solution of the diffusion equation.

10. Using variables separable method, the only solution for the given PDE, which isphysically acceptable is of the form

2

1 2( , ) ( cos sin ) tx t c x c x e ��

�

� � (1)

Applying the BC: 0, 0,x �� Eq. (1) becomes

2

1( , ) cos tx t c xe ��

�

�(2)

Applying the second BC: , 0,x a θ= = we find 1 0.c �

Therefore, the possible solution is

2

2( , ) sin 0 for allta t c e tλθ λα −= =

Since 2 0, sin 0,c a�� implies ,a n� �� therefore, / ,n aλ π= where n is any integer.

Hence, the possible solution is

2 2

2( , ) sin

nt

an

n xx t b e

a

ππθ

−⎛ ⎞= ⎜ ⎟⎝ ⎠

(3)

where 2.nb c� Adding all possible solutions the most general solution satisfying thegiven BCs: is

2 2

21

( , ) sin expnn

n x nx t b t

a a

� �

�

�

�

� �� (4)

Finally, we have to satisfy the IC: 0( ,0)x� �� (constant). Therefore,

01

sin ,nn

n xb

a

��

�

� ��

� which is a half range



Fourier sine series and hence, the required solution is given by Eq. (4) where nb isgiven by

00

2sin

a

nn x

b dxa a

πθ ⎛ ⎞= ⎜ ⎟⎝ ⎠∫ (v)

��

1. From D’Alembert’s method, the required solution is of the form

1( , ) [ ( ) ( )]

2u x t f x ct f x ct� � � �

Given

0( ) sinx

u f x ul

��

it follows that

0 ( ) ( )( , ) sin sin

2

u x ct x ctu x t

l l

� ��

�

Using

1sin cos [sin ( ) sin ( )]

2� � � � � ��

we have

0( , ) sin cosx ct

u x t ul l

� ��

2. The general solution is given by Eq. (4.37). Using the IC: ( , 0) 0,tu x � we have

0.nB � The use of another IC: ( ,0) 10 sin ( / )u x x lπ= yields

10 sin sinnx n x

Al l

π π=∑implying thereby 1 2 310, 0.A A A� � � �� Hence the required solution is

( , ) 10 cos sinct x

u x tl l

� ��

3. The general solution is given by Eq. (4.35):

1 40, 0 0, 0 gives 0tu x c u c= = ⇒ = = =



The solution reduces to: 3( , ) sin cos . 0u x t c x c t uλ λ= = at .x n� �� Thus thegeneral solution is

1

( , ) sin cos ( )nn

u x t b nx nct�

��

where

1

0

2 2sin ( 1)n

nb x nx dxn

π

π+= = −∫

4.3 3

( , ) 9sin sin sin sin12

bl x c t x c tu x t

c l l l l

� � � �

�

� ��

� �

5. From Eq. (4.37), the possible solution is

( , ) sin cos sinn nn x n n

u x t A ct B ctl l l

π π π⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦∑

Applying the IC: ( ,0) 0,tu x � the possible solution is of the form

1

( , ) cos sin cos sinn nn

n c n ct n x n ct n xu x t B B

l l l l l

� � � � �

�

� � ��

�

(i)

To find the coefficients nB� we use the IC: ( ,0) ( )u x f x= to get

1

( ) sinnn

n xf x B

l

�

�

��

where

0

2( ) sin

l

nn x

B f x dxl l

π′ = ∫This form of the solution does not convey the fact that the initial disturbance will bepropagated as some sort of wave motion. But if we rewrite equation (i) in thealternative form

1

1( , ) sin ( ) sin ( ,

2 nn

n nu x t B x ct x ct

l l

� �

�

� ��

��

it gives a feeling that ( , )u x t is a sum of two travelling waves, one moving to theright and the other moving to the left with speed c.

6. Since u is a function of r and t only, the given equation reduces to

2 2

2 2 2

2 1u u u

r rr c t+ =� � �

��



Let / ,u rφ= the governing equation further simplifies to the form

2 2

2 2 2

1

r c t

φ φ=� ��

the solution of which is

( ) ( )f r ct g r ctφ = − + +Hence the general solution of the given equation is

[ ( ) ( )]/u f r ct g r ct r= − + +

7. 2 2 32

1 1( , ) 5 ( 2 ).

3 2x ct x ct xu x t x t c t e e e

t� �

� � � � � �

8. ( , ) sin cos ( 1) ( ) .t tu x t x ct e xt x xte� � � � �

9. We notice that one boundary condition is a function of time. Using D’Alembertsmethod, the solution of the wave equation is ( , ) ( ) ( ).u x t x ct x ct� �� The boundary

conditions give ( ) ( ) 0x xψ φ+ = and ( ) ( )x x A� �� (the constant of integration).

Their solution gives ( ) / 2,x A� � ( ) / 2x A� � � when their arguments are positive.

Now applying the IC, we get ( ) sin ( ).ct t ct� �� But for 0, ( ) /2.t ct Aψ> =Therefore, ( ) sin /2ct t Aφ − = − for 0.t � Now, when 0x ct� � or / ,t x c> we have

( ) sin /2x x

x ct c t t Ac c

φ φ ⎡ ⎤⎛ ⎞ ⎛ ⎞− = − − = − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Combining these results, we get

0 when

( , )sin when

x ct

u x t xt x ct

c

��

10. Similar to Example 4.9 with 1, y� �� etc.

11. The vibration of the string is described by

2PDE: , 0 , 0tt xxy c y x L t= ≤ ≤ >

where 2 / ,c τ ρ= subject to

BCs: (0, ) ( , ) 0y t y L t= = for all t and

ICs: ( , 0) 0ty x = .

The initial displacement of the string is given by

2 / , 0 /2( , 0)

2 ( )/ , / 2

sx L x Ly x

s L x L L x L

< ≤⎧⎪= ⎨− ≤ ≤⎪⎩



Using variables separable method, the displacement of the string at any time is givenby

21

8( , ) sin cos sin

2( )n

s n n ct n xy x t

L Ln

� � �

��

� � � � � ��

�

where 2 / .c τ ρ=or

2 2 2 2

8 1 1 3 3 1 5 5( , ) sin cos sin cos sin cos

3 5

s x ct x ct x cty x t

L L L L L Ll

π π π π π ππ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − × + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦�

12. The first term in the solution of problem 11 is the normal mode corresponding to thelowest normal frequency often called Fundamental frequency. That is,

2

8sin cos

s x ct

L L

� �

�

� � � ��

.

Let 1f be the corresponding frequency, then

1 11

2 / or /2 /2

f c L f c LL

π π τ ρ= = =

Since 1 cos 1,� � � the mode is such that, the string oscillates as shown in Fig. 4.9(a)from the heavy to the dotted curve and back to the heavy curve and so on.

Lx

y

O

Fig. 4.9(a) First normal mode.

The next higher frequency is given by the mode which corresponds to the term

2

8 3 3sin cos

9

s x ct

L L

� �

�

� � � ��

Let 3f be the corresponding frequency, then

3 33 3

2 3 / or /2 2

cf c L f

L Lπ π τ ρ= = =



The mode in this case is depicted in Fig. 4.9(b).

Lx

y

O

Fig. 4.9(b) The third normal mode.

Similarly, we find the higher normal frequencies

5 75 7

/ , /2 2

f fL L

τ ρ τ ρ= = and so on.

However, we observe that the even frequencies 2 4,f f etc. are absent. But, it shouldbe noted that both even and odd frequencies will be present in a general displacement.

13. The governing PDE: 2 , 0tt xxy c y x L� � � subject to the BCs:

00, ( , ) 0

x

yy L t

x == =�

�and the ICs:

00( ,0) cos ( /2 ), 0t

yy x y x L

tπ

== =�

�The time-dependent displacement of the string is found to be

0( , ) cos sin .2 2

x c ty x t y

L L

� ��

14. Let ( ) ( )u f x vt i y g x vt i y� �� .

Then,

,xu f g′ ′= + xxu f g�� (1)

,yu i f i g� �� 2 2

yyu f gα α′′ ′′= − − (2)

,tf v f v g� �� 2 2

ttf v f v g′′ ′′= + (3)

Now, substituting expressions (1), (2) and (3) in the given PDE, we find that

22 2

2( )

vf g f g f g

c� ��



or2

2 22

(1 ) (1 ) ( )v

f g f gc

� ��

is satisfied, provided2 2 2 2 2 21 / or 1 / .v c v cα α− = = −

15. The D’Alembert’s solution for the IVP defined as

2PDE: , 0,tt xxu c u t x� � ��

ICs: ( ,0) ( ), ( ,0) ( )tu x x u x v x�� is

1 1( , ) [ ( ) ( )] ( )

2 2

x ct


cη η ξ ξ

+

−= + + − = ∫

In the given problem, ( ) 0, 2 and ( ) .v x c x x�� Therefore, the solution to the givenproblem is

1 1( , ) [ ( 2 ) ( 2 )] [ ]

2 2u x t x t x t x ct x ct x� ��

Hence, the correct choice is (A).16. D’Alembert’s solution is given as

1 1

( , ) [ ( ) ( )] cos2 2

x t

x tu x t x t x t dη η ξ ξ

+

−= + + − + ∫

1 1[sin ( ) sin ( )] [sin ( ) sin ( )] sin ( )

2 2x t x t x t x t x t� � � � � � � � � �

Therefore,

2, sin sin sin 120 3/2

2 6 2 6 3u

π π π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Hence, the correct choice is (A).

17.2 ( )

0( , ) ( ) sin

tn t

nu x t a e nx d�� Chapter 5

1. Let ( , , ; , , )x y zδ ξ η ζ be a Dirac -functionδ defined by the equation

( , , ) ( , , ; , , ) ( , , )f x y z d d d f x y z� � � � � � � � � ��

��



which is valid for all continuous functions ( , , )f x y z having compact support in the-( , , ) space.x y z If ( ), ( ), ( )x y z� � � � � �� are three one-dimensional -functions,δ

then we have

( , , ) ( ) ( ) ( ) ( , , )f x y z d d d f x y z� � � � � � � � ��

��

Hence comparing the above two equations, we get

( , , ; , , ) ( ) ( ) ( )x y z x y zδ ξ η ζ δ ξ δ η δ ζ= − − −

2. Suppose we transform from cartesian coordinates x, y to curvilinear coordinates ,ξ ηsatisfying the realtions ( , ); ( , ),x f y gξ η ξ η= = where f and g are single-valued

continuously differentiable functions. Also suppose that 1 2, ,� � � �� corresponding

to 1 2, .x y� �� Then the equation

1 2 1 2( , ) ( ) ( ) ( , )x y x y dx dy� � ��

becomes

1 2 1 2( , ) [ ( , ) ] [ ( , ) ] | | ( , )f g f g J d dφ δ ξ η α δ ξ η α ξ η φ α α− − =∫∫where

( , )

( , )

f f f gJ

g gη

ξ η

ξ ξ η= = �

�

The above equation indicates

1 2 1 2 1 2[ ( , ) ] [ ( , ) ] | | ( ) ( ) | | ( ) ( )f g J x y Jδ ξ η α δ ξ η α δ α δ α δ ξ β δ η β− − = − − = − −

In polar coordinates, let cos , sin .x r y r� �� Then ( , ), ( , ).x f r y g r� �� Thus,

cos sin| |

sin cosr

r

f f rJ r

g g r�

�

� �

� �

�

� � �

Hence

0 00 0

( ) ( )( ) ( )

r rx x y y

r

� � � ��

� �

� � �

3. Let sin cos , sin sin , cos .x r y r z r� � � � �� That is, x = f(r, �, �), y = g(r, �, �),

( , ).z h r �� Then

2| | sin

r

r

r

f f f

J g g g r

h h h

� �

� �

� �

��



It therefore, follows that

1 1 11 1 1 2

( ) ( ) ( )( ) ( ) ( )

sin

r rx x y y z z

r

� � � � � � ��

�

� � �

� � � �

5. 2 2

( )( , ) .

( )

y f x dxu x y

x x y�

�

��

� ��

6. In the two-dimensional case, we can place singularity at (x, y) and consider its imageover one side, again reflect the image over the image of that side, and so on untilwe return, after an even number of reflections to the original domain. Let ( , )x y′ ′− be

the image point of ( , )x y� � in the y-axis and let ( , ), ( , )x y x y� � � �� be the image ofthe above two points in the x-axis. Then G may be constructed by means of threesuitably placed unit charges, a positive charge at ( , )x y′ ′− and negative charges at

( , ) and ( , ).x y x y� � � �� Thus

2 2 2 2

2 2 2

1 {( ) ( ) }{( ) ( ) }( , ) ln

4 {( ) ( ) }{( ) ( ) }

x x y y x x y yG x y

x x y y x x y yπ⎡ ⎤′ ′ ′ ′− + − − + += ⎢ ⎥

′ ′ ′ ′+ + − + + +⎢ ⎥⎣ ⎦

This function satisfies 2 0G∇ = except at the source point ( , )x y� � and G = 0 on

x = 0; also, / 0G n =� � on y = 0.

��

1. (i) 2 2 2 2 2 2 2 2 2( ) /( ) ; (ii) 2( 1) /( 2 1) ; (iii) 4( 1) /( 2 5) ;s a s a s s s s s s� � � � � � � �

(iv) 2 3

1 2 3.

( ) ( ) ( )s a s a s a� �

� � �

2. (i) 2(1 ) / .se sτ τ−−

3. (i) 2 11 1ln ; (ii) [ln ( 4)] [tan ( / 3)] .

2 s ss

s ss

∞ − ∞−⎛ ⎞ + −⎜ ⎟⎝ ⎠

4.2 2

1 1( 1)

1

ss

s

ee

se s

�

�

�

� ��

� � ��

5. 6 2 6(1 ) / (1 )s se s e− −− +

6. (1 ) / (1 )bs bse s e− −− +

7. 2 se s−



8. (i) 22 3/2

1(1 / 1); (ii)

( 1)s s

s− +

+

9. (i) 21 1;

2 2t te e� �

� � (ii) 2 2( );te t t� (iii) 1 cos ;t�

(iv) 21 13 ;

3 3t t te te e�� (v)

3

[ sin 2 ];4

tet t

�

(vi) 21 52

2 2t te e� �

� �

10. (i) ( )/ ;bt ate e t− −− (ii)sin

;kt

t�

(iii) 2 2cos;

te t

t

� (iv)42cos 2

.tt e

t

�

12. (i)3

sin;

at at

a

� (ii) ( );ate

erf ata

(iii) (sin 2 2 cos 2 ).64

tt t t�

13. 1 2 0,[ /( 1)]

sin ,s t

L e st t

��

�

� �

��

� ��

14. (i) /21 3 33sin cos ;

3 2 2t t t t

e e� �

� ��

� � �

(ii) 2 sin 2 coste t t� �

15.2 2( 1/ 2)

1

4 ( 1)1 cos [( 1/ 2) ]

(2 1)

nn t

n

n x en

��

�

��

�

��

� 16. 2 27 4 4t t ty e e te� � � �

17.1

(cos cosh )2

y kt kt� �

18.2/2ty e−=

19. 1, 2t tx e y e� �

� � � �

20. 3 sin cos ;t tx e t e t� �

� �

[2 cos 1 sin ]ty e t t�

� � �



21.cos cos

, 14 4 2 4 4 2

t t t te e t e e tx y

� �

� � � � � � �

22. The Laplace transform of the given PDE yields

2

2

d H sH

kdx�

The Laplace transform of the BCs gives

1when 0

( 1)H x

s s� �

�

0 whenH x ��

Hence the solution of the above ODE is

sin h ( )1

( 1)sin h

sx

kH

s s sx

k

�

� ��

� ��

� � �

Taking the inverse Laplace transform using inversion formula, we get

sin h ( )1

( , )2

( 1) sin h

st

i

i

se x

kx t ds

i ss s

k

�

�

�

��

�

� �

� �

� ��

��

� � ��

�

The integrand has simple poles at 20, 1 and , 1, 2,s n k n= − − = …23. After taking the Laplace transform of the given PDE and using the IC, we obtain

2

2( , ) 10 cos 5

3

d U sU x s x

dx� � �

Its general solution is

/ 3 30cos5( , ) exp

3 75s x s x

U x s Ae B xs

⎛ ⎞= + − +⎜ ⎟ +⎝ ⎠

Taking the Laplace transform of the BCs and utilizing them into the above solution,we have only the trivial solution 0.A B� �

Thus we get

30cos5( , )

75

xU x s

s�

�



its inverse Laplace transform yields

1 7530( , ) cos 5 ; 30 cos 5

75tu x t L x t e x

s� �

� ��

24. Taking the Laplace transform of the given PDE and using the ICs, we obtain

2 2

2 2( , ) 0

d U sU x s

dx c� �

Its general solution is

( , ) cosh sinhs s

U x s A x B xc c

� � � ��

Taking the Laplace transform of the BCs, we get

(0, ) 0, ( , )xP

U s U l sEs

= =

Using these expressions, the general solution reduces to

2( , ) sinh

cosh

PC sU x s x

s cEs l

c

� ��

Taking inverse Laplace transform and using complex inversion formula, we get

2

1( , ) sinh

2cosh

sti

i

PCe su x t x ds

si cEs l

c

�

��

� �

� �

� ��

�

22

0

8( 1) (2 1) sin ( ) cos ( )n

n nn

Pl xn s x s ct

E l

��

�

� ��

� ��

where

2 1

2nn

sl

��

�

25. Taking the Laplace transform of the given PDE, we obtain

22 2

2( , ) ( ,0) ( ,0)t

d Us U x s su x u x c

dx� � �

Using the initial conditions, we get

2 2

2 2 2

1d U sU

dx c c� � �




2

1( , ) exp exp

s sU x s A x B x

c c s

� � � ��

As ,x � the transform should be bounded, which is possible only if 0.B � Also,

using the Laplace transform of the BC, i.e., (0, ) 0,U s � we get 21/ .A s= − Hence,

2 2

1 1( , ) exp

sU x s x

cs s

� ��

Taking the inverse Laplace transform of the above equation, we obtain

( , )x x

u x t t t H tc c

⎡ ⎤⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

where H is the Heaviside unit function.

26., if

( , )

0, if

x xf t t

a ax tx

ta

�

� � ��

where 1/a LC� .

��

1.21 12

1 1

2

3

1 1 1( ) (1 ) ( )

2 2

2 22

2 22

2 cos sin2

i x i x

i i i i

xF x e dx d e

i

e e e e

i

α α

α α α α

απ π α

απα

α α απ α

− −

− −

−= − =

⎡ ⎤⎛ ⎞ ⎛ ⎞+ −= − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

−⎛ ⎞= − ⎜ ⎟⎝ ⎠

∫ ∫

Then,

3

2

1 2 cos sin( ) 2

2

1 , | | 1

0, | | 1

i xf x e dx

x x

x

αα α απ π α

� ��

��

� � ��

�



Therefore,

2

3

(1 ), | | 1cos sincos 2

0, | | 1

x xx d

x

πα α α α α

α

�

��

� � ��

�

Set 1

;2

x � we then have

3

cos sin 1 3cos 1

2 2 4 8d

α α α α πα πα

�

��

� � �� Therefore,

30

cos sin 3cos

2 6

x x x xdx

x

π� � � ��

2.20

20

2 2( ) sin

1

2 1 2( ) cos

1

xs

xc

F e x dx

F e x dx

��

� ��

�

�

�

�

�

�

� ��

� ��

3.2

( )cF e ��

�

�

�

4. From Examples 7.3 and 7.4, we have

2 20

2 2( ) cosax

cG e x dxa

αα απ π α

� ��

Also,

0 0

2 2( ) ( )cos cos

2 sin

cF f x x dx x dx

x

λα α α

π π

απ α

��

�

� �

using Parseval’s relation

0 0( ) ( ) ax

c cF G d e dxλ

α α α� ��



or

2 20 0

2 sin

( )axa

d e dxa

λαλ απ α α

� ��

Therefore,

2 2 20

sin 1

2( )

aed

a a

λαλ παα α

�� 7. Taking the Fourier transform of PDE, we get

2 1( )

2

dUU t

dt� �

��

After applying the transform of IC

1( , 0)

2U �

�

�

its solution21

( , )2

tU t e ��

�

�

�

Taking the inverse transform, of this result, we obtain

2 2/41 1 1( , )

2 2t i x x tu x t e e d e

tα α α

π π� � � ��

� ��8. Duhamel’s principle is: The solution of the homogeneous PDE with a non-homogeneous

initial condition is equivalent to solving non-homogeneous PDE with a homogeneousinitial condition.In fact, from Example 7.13, the solution of the initial value problem

0, , 0

( , 0) ( ),

t xxv v x t

v x f x x

� � ��

� ��

is

21 ( )( , ) ( ) exp

4 4

xv x t f d

t t

�

� �

�

�

��

� ��

� ��

Now, applying Duhamel’s principle, the solution of the non-homogeneous problem

( ), , 0

( , 0) 0,

t xxv v f x x t

v x x

� � ��

� ��



is

2

0 0

1 ( )( , ) ( , ) ( ) exp

4 ( ) 4( )

t t xv x t v x t d f d d

t t

ατ τ α α τπ τ τ

�

��

� ��

Therefore, the Green’s function for the non-homogeneous problem is

21 ( )( , ; , ) exp , 0

4 ( ) 4( )

xG x t t

t t

��

� � �

� ��

� ��

11. From Example 7.18, the solution of Laplace equation when ( , 0) ( )u x f x� is givenin Eq. (7.164). Therefore, in the present problem

0, 0( )

1, 0

xf x

x

��

��

the required solution is given by

0

2 20

1( , ) 0

( )

y du x y d

x y

ξξπ ξ

�

��

� ��

or

1 1 1 1

0

1 1 1( , ) tan tan tan tan

2

y x x xu x y

y y y y

ξ ππ π π

��

12. Finally, taking finite cosine Fourier transform, we have, after inversion, the relation

2 2

2 21

24 cos 1( , ) 6 exp cos

36 6n

n n t n xV x t

n

π π ππ

�

�

� ��

13.01

2 2 ( 1) cosh( , ) sin .

cosh

n

n

nyV x y nx

n nyπ π

�

�

��


Absolute convergence, 392Adjoint operator, 69, 70

to Laplace operator, 71self, 71, 73

Amplitude, 238Angular frequency, 238Auxiliary equation of PDE, 13, 14, 17Axisymmetric fluid flow, 172

Besselequation, 140, 154, 173, 261, 263functions, 82, 140, 141, 144, 213, 265

Biharmonic PDE, 169Boundary value problems, 109, 419Burger equation, 52, 218, 219

Canonical form of PDE, 53, 55–60, 236Cauchy problem, 21, 73, 75

first order equation, 21second order equation, 103

Cauchy method of characteristics, 25, 29Characteristic(s),

curves, 13, 26, 28equations of PDE, 29, 30strip, 29

Charpit’sequations, 38method, 37

Churchill’s problem, 110Circular frequency, 238Clairaut’s form, 46, 47Classification of PDE, 53Compatibility condition, 29Complete integral/solution, 38, 39, 41Continuity equation, 155Convolution theorem, 344, 412Current density, 235

D’Alembert’s solution, 240, 426Diffusion equation, 182, 185, 206, 208, 211

in cylindrical coordinates, 208fundamental solution of, 186in spherical polar coordinates, 211one-dimensional, 204singularity solution of, 287two-dimensional, 206

Dirac delta function, 189, 191, 282, 314two-dimensional, 192

Direction cosines, 4, 6, 11Dirichlet condition, 184, 257, 388

exterior problem, 109interior problem, 109

Discriminant, 53Divergence theorem, 110, 155, 183, 286, 292Domain of dependence, 242Duhamel’s principle, 268

for heat equation, 483for wave equation, 268

485

Index


486 INDEX

Eigenfunction method, 257, 302Eigenfunctions, 258, 302, 303Eigenvalues, 258, 302, 303Electric

charge density, 235field, 235potential, 151

Electrostatic potential, 286Elementary solutions, 185

of heat equation, 185Elliptic PDE, 106Error function, 188, 194, 332

its complementary, 194, 333Euler’s equation, 147Euler-Ostrogradsky equation, 234Euler-Poisson-Darbaux equation, 279Exponential order, 317

Faultung theorem, 344, 412First order PDE, 1

linear, 2non-linear, 2, 23quasilinear, 1

Flexible string, 233Flow, 172

in a finite circular pipe, 172in a semi-infinite circular pipe, 172past a cylinder, 169

Force function, 269Fourier–Bessel

function, 141series, 141orthogonality property, 141, 144

Fourier coefficients, 389Fourier integral, 390

cosine, 394cosine convolution, 413exponential form of, 395formula, 426sine convolution, 413theorem, 395

Fourier law, 182Fourier series, 389

double, 208generalised, 151sine, 136

Fourier transform, 388, 395cosine, 394of Dirac delta function, 416of elementary function, 396inverse, 396multiple, 416pairs, 396properties of, 401sine, 396

Frequency, 239angular, 238

Gamma function, 321Gauss divergence theorem, 110Gauss law, 108General integral, 14General solution

of first order PDE, 16Gravitational potential, 106Green’s function, 82, 228, 282

for diffusion equation, 310for heat flow in finite rod, 313for heat flow in infinite rod, 313by the method of images, 295properties of, 291

Green’s identities, 111Green’s theorem, 73

in a plane, 243

Hadamard’s example, 180Hamilton’s principle, 234Hankel’s function, 261Harmonic function, 111

maximum-minimum principle of, 115mean value theorem for, 114

Heat equation, 182in plate, 206in rods, 201

Heat flowin an infinite medium, 421in a semi-infinite medium, 422

Heat flux, 182Heaviside expansion theorem, 350Heaviside unit step function, 283, 349Heine-Borel theorem, 116


INDEX 487

Helix, 6Helmholtz theorem, 305Helmholtz equation, 305Hopf-Cole function, 220Hopf-Cole transformation, 218Hyperbolic PDE, 232

IBVP, 189Impulsive force, 189Integral surface of PDE, 12, 14, 18, 20, 24, 31Integral curves, 13, 23Irrotational flow, 154

of incompressible fluid, 154IVP, 275

Jacobian, 3, 53

Kernel, 282Kinetic energy, 234, 251

Lagrange’s identity, 70, 73Lagrange’s method, 16Laplace equation, 106

in cartesian form, 118in cylindrical form, 118in plane polar form, 118in spherical polar form, 120

Laplace equation solutionin an annulus, 164in axisymmetric case, 149in a rectangular box, 158in a rectangular plate, 156in a sphere, 151between two concentric spheres, 161

Laplace transform, 316of Bessel function, 335convolution theorem, 344of Dirac delta function, 337of error function, 332initial value theorem, 326inverse, 337, 338of a periodic function, 329properties of, 321–323of unit step function, 349

Legendre, 213associated, 213equation, 148function, 148polynomial, 151

Leibnitz rule, 268Level surface, 3Limit of a sequence, 192L’Hospital rule, 317, 351

Magnetic field, 235Maximum-minimum principle, 115, 215

and consequences, 115, 215for heat equation, 215for Laplace equation, 115

Maxwell’s equation, 235Mean value theorem, 114

for harmonic function, 114of integral calculus, 190, 191second, 391

Mellin–Fourier integral, 352Method of images, 295Method of residues, 364Method of variables separable, 122Modes of vibration, 247, 280

first normal mode, 248third normal mode, 248

Monge cone, 26direction, 26

Neumann, 110condition, 110, 184, 257

Newton’s law of cooling, 185Non-linear equations, 217, 218Normal density function, 431Normal modes of vibration, 247, 280

Order of PDE, 52Orthogonality property, 144, 163

of Bessel’s function, 141, 144of eigenfunction, 226, 258of Legendre polynomial, 151, 163, 169, 215,

255of sine function, 255


488 INDEX

Partial Differential Equations (PDE)biharmonic, 169canonical form of, 53, 55, 57, 59classification of, 53elliptic, 106of first order, 1, 11hyperbolic, 232of second order, 52, 182, 232parabolic, 182

Poisson equation, 108Poisson integral formula, 132, 300Potential, gravitational, 143, 151Potential energy, 234, 250, 251Pulse function, 269

Recurrence relation, 145, 146, 432Reimann

function, 74Green’s function, 74Lebesgue lemma, 390localisation lemma, 391method, 71Volterra solution, 244

Robin’s condition, 185, 257

Singularity, 148Singularity solution, 282

of diffusion equation, 287of Helmholtz equation, 305of Laplace equation, 287

Specific heat, 182Spherical

coordinates, 120mean, 113symmetry, 153

Stability theorem, 116Stokes flow, 169

Superposition principle, 127, 130, 141, 197, 200,247, 251, 265

Surfaces, 2

Tangent plane, 5Test function, 283Thermal conductivity, 182Thermal diffusivity, 184Torsion of a beam, 180Tricomi equation, 63

Uniqueness solution ofDirichlet problem, 112heat equation, 216wave equation, 266

Unit impulse function, 190

Variables separable method, 122, 126, 132, 136,138, 146, 151, 170, 177, 195, 245

Vibration, 245forced, 254of a circular membrane, 264of parameter method, 255of a rectangular membrane, 274transverse, 245, 425

Wave equation, 232in cylindrical polar, 260D’Alembert’s solution, 240, 241Green’s function for, 305Riemann-Volterra solution, 244in spherical polar, 262three-dimensional, 52unique solution, 266

Wave length, 239Wave number, 239Well-posed problem, 181


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